Answer:
120.237 seconds
Explanation:
Given that:
V = 144 V
I = 10.5 A
H = 60.6 kJ
Using the formula:
H = I²RT
From H = I²RT; making T the subject, we have:
[tex]T = \dfrac{H}{I^2R}[/tex]
where;
[tex]R = \dfrac{V}{I}[/tex]
∴
[tex]T = \dfrac{H}{V \times I}[/tex]
[tex]T = \dfrac{60.6 \times 10^3 }{144 \times 10.5}[/tex]
T = 40.079
[tex]T_{neq} = 3T[/tex]
[tex]T_{neq} =3 \times 40.079[/tex]
[tex]\mathbf{T_{neq} =120.237 \ sec}[/tex]
6. The petrol in a petrol can weighs 2000g. The density of petrol is 0.8g/cm3.
What is the volume of the petrol in the can in a) cm3 b)litres (1000cm3=1 litre)
Pls help :((
Answer:
a. 2500 cm³.
b. 2.5 litres.
Explanation:
Given the following data:
Density = 0.8g/cm³
Mass = 2000g
To find the volume of the petrol;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
[tex]Density = \frac{mass}{volume}[/tex]
Making volume the subject of formula, we have;
[tex]Volume = \frac{mass}{density}[/tex]
Substituting into the equation, we have:
[tex]Volume = \frac{2000}{0.8}[/tex]
Volume = 2500 cm³
a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.
b. The volume of the petrol in the can in litres;
1000 cm³ = 1 litre
2500 cm³ = x litres
Cross-multiplying, we have;
1000x = 2500
x = 2500/1000
x = 2.5 litres.
Therefore, the volume of the petrol in the can in litres is 2.5.
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreciable air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. Part A What is the maximum height this rocket will reach above the launch pad
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
In order for the eye to see an object _____ from the object myst be reflected to your eye.
Light or particle ?
Answer: light from the object
Explanation:
When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light
6th grade science I mark as brainliest !
Answer:
first is Atoms
4) is True
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is launched vertically into the air at point T. In case B, a 1 kg block slides without friction down an identically shaped ramp and is also launched vertically at point T. Select the statement that best describes which object will go higher after launch, and why
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = [tex]\frac{2}{5}[/tex] m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²
m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
[tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
Which of the following variables can be measured in joules?
A. momentum
B. Energy
C. Power
D. Work
Answer:
The variables that can be measured in joules are
B. Energy
D. Work
Hope it will help :)
The number of complete wavelengths that pass a point in a given time is referred to as...
A. Wavelength
B. Frequency
C. Amplitude
D. Reflection
In the 1986 Olympic Games, Abdon Pamich of Italy won the 50 km walk, in
4h, 11 min and 11.2 s. Find his average speed in m/s.
Answer: 4
Explanation: because my pet monkey said it was
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?
Answer:
it 1.5 meters
Explanation:
if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)
The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If the maximum speed is 35 mi/hr for a 3000-lb car, compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Answer:
the magnitude F of the total friction force is 2456.7 lb
Explanation:
Given the data in the question;
maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s
diameter = 200ft
radius = 200/2 = 100 ft
First we calculate the normal component of the acceleration;
[tex]a_{n}[/tex] = v² / p
where v is the velocity of the car( 51.3333 ft/s)
p is the radius of the curvature( 100 ft)
so we substitute
[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft
[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft
[tex]a_{n}[/tex] = 26.35 ft/s²
we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)
1 ft/s² = 0.0310809502 g
[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g
[tex]a_{n}[/tex] = 0.8189g
Now consider the dynamic equilibrium of forces in the Normal Direction;
∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]
F = m[tex]a_{n}[/tex]
we know that mass of the car is 3000-lb = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)
so
we substitute
F = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb) × 0.8189g
F = 2456.7 lb
Therefore; the magnitude F of the total friction force is 2456.7 lb
A laser positioned on a ship is used to communicate with a small two man research submarine resting on the bottom of a lake. The laser is positioned 12 m above the surface of the water, and it strikes the water 20 m from the side of the ship. The water is 76 m deep and has an index of refraction of 1.33. How far is the submarine from the side of the ship
Answer:
84.1 m
Explanation:
Given :
The distance from the ship to submarine :
20 + y
Using Pythagoras :
Tan θ = opposite / Adjacent
Tan θ = 20 / 12
12 tan θ = 20
θ = tan^-1(20/12)
20
θ = 59.036°
The angle phi;
n1sinθ1 = n2sin θ
Sin 59.036 = 1.33 * sin phi
Sin phi = sinsin(59.04) ÷1.33
0.8574907 = 1.33 * sin phi
Sin phi = 0.8574907 / 1.33
Sin phi = 0.6447298
phi = sin(0.6447298
Phi = 40.15°
From Pythagoras :
y = 76tan40.15°
y = 76 * 0.8435707
y = 64.11
20 + y
20 + 64.11 = 84.11
Suppose a car is traveling at 22.8 m/s, and the driver sees a traffic light turn red. After 0.404 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 9.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?
Answer:
38.09 m
Explanation:
We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:
Reaction time (tᵣ) = 0.404 s
Initial velocity (u) = 22.8 m/s,
Distance travelled during the reaction time (sᵣ) =?
sᵣ = utᵣ
sᵣ = 22.8 × 0.404
sᵣ = 9.21 m
Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:
Initial velocity (u) = 22.8 m/s
Acceleration (a) = –9 m/s² (since the car is decelerating)
Final velocity (v) = 0 m/s
Distance travelled when the brake was applied (s₆) =?
v² = u² + 2as₆
0² = 22.8² + (2 × –9 × s₆)
0 = 519.84 – 18s₆
Collect like terms
0 – 519.84 = –18s₆
–519.84 = –18s₆
Divide both side by –18
s₆ = –519.84 / –18
s₆ = 28.88 m
Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:
Distance travelled during the reaction time (sᵣ) = 9.21 m
Distance travelled when the brake was applied (s₆) = 28.88 m
Stopping distance =?
Stopping distance = sᵣ + s₆
Stopping distance = 9.21 + 28.88
Stopping distance = 38.09 m
The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
A football of mass 2.5kg is lifted up to the top of a cliff that is 180m high. How much
potential energy does the football gain?
The potential energy of the football with mass 2.5 kg which is lifted up to the top of a cliff is 4410 Joules.
What is Potential energy?Potential energy is the stored energy which depends upon the relative position of the various parts of a system of objects. Potential energy is the product of mass of the object, acceleration due to gravity, and the height. The SI unit of potential energy is Joule (J).
PE = m × g × h
PE = Potential energy,
m = mass of the object,
g = acceleration due to gravity,
h = height
PE = 2.5 × 9.8 × 180
PE = 4410 Joules
Therefore, the potential energy of the football is 4410 Joules.
Learn more about Potential energy here:
https://brainly.com/question/24284560
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The design of interior spaces is relatively unimportant to good
architecture?
Could I get help on this question please
Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
Based on your graph, predict the length of pendulum that will give you a
period of 2.00s.
Answer:
About 1 meter
Explanation:
Base on your image, when the period is 2 (horizontal variable), the length of the pendulum is about 1 meter.
A place kicker must kick the football from a point [06] m from the goal and clear a bar 3.00 m above the ground. The ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0 degrees above the horizontal. (a) By how much (m) does the ball clear (positive value) or fall short (negative value) of the cross bar? (This is vertical distance above or below the cross bar.) (b) When it gets to the cross bar, what is the vertical component of the ball’s velocity (m/s)? (Is it rising or falling-pay attention to the sign?)
Answer:
Explanation:
Horizontal component of initial velocity of throw = 20 cos 53 = 12 m /s
Vertical component = 20 sin 53 = 15.97 m /s
Distance to be travelled horizontally = 6 m .
time taken by ball to travel this distance = 6 / 12 = 0.5 s
vertical displacement during this period can be calculated as follows .
Initial vertical velocity = 15.97 m /s
time of travel = .5 s
acceleration = - 9.8 m /s²
s = ut - 1/2 g t²
= 15.97 x .5 - .5 x 9.8 x 0.5²
= 7.985 - 1.225
= 6.76 m
Goal post is 6 m high , so ball will cross the goal post .76 m or 76 cm above cross bar .
b ) vertical component of ball when it crosses the goal post . Let it be v .
v = u - gt
Applying this formula for vertical movement ,
v = 15.97 - 9.8 x .5
= 15.97 - 4.9
= 11.07 m /s .
A lunar eclipse occurs when the Moon passes through Earth's
An electric charge at rest produce
Answer:
Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.
plz follow me
the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
[tex]W_{net}[/tex] = m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
[tex]W_{net}[/tex] = m (gh)
m = [tex]W_{net}[/tex] / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
WILL GIVE BRAINLIEST!!
What is a way to transfer charge in which an object becomes polarized?
Answer:
I answered Number 4 (Solids and Elasticity)
Explanation:
solids and elasticity
If a skaters mass increases how does that effect kinetic energy
Answer:
By paying close attention to the formula for average kinetic energy, we can see that by increasing the mass by a proportional amount will lead to an increase in the total average kinetic energy. There is a direct relationship being observed between the values.
A baseball player hits a baseball. The mass of the ball is 0.15 kg. The ball accelerates at a rate of 60 m/s 2 . What is the net force on the ball to the nearest newton?
Answer:
Please find attached pdf
Explanation:
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
Your favorite golfer taps the golf ball with just enough force that it rolls into the ninth hole is an example of what law of motion???
Answer:
mass and on the net force acting on it. ... Tap again to see term ... Newton's second law of motion states that an object's acceleration depends on its ... You hit a ping-pong ball & a tennis ball with a tennis rack
A swimmer pushing off from the wall of a pool exerts a force of 1 newton on the wall. What is the reaction force of the wall on the swimmer?
Answer: 1 Newton
Explanation:
"Every action has an equal and opposite reaction."
Please mark as Brainliest if it is correct.
Force is an action-reaction principle. It stated that the force always exists in a pair. The reaction force of the wall on the swimmer will be 1N.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Both occur in an action-reaction form. Hence for every action, there is an equal and opposite reaction.
[tex]\rm F_{action}=F_{reaction} \\\\ \rm F_{action= 1N[/tex]
[tex]\rm F_{reaction} = 1N[/tex]
Hence the reaction force of the wall on the swimmer will be 1N.
To learn more about Newton's third law refer to the link;
https://brainly.com/question/1077877
6.
ribbon
AA
SON
120 N
Two teams of students are competing in a tug-o-war contest, as shown in the
picture above. How does the ribbon move?
Answer:
The ribbon will move to the right.
Explanation:
To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:
Force to the right (Fᵣ) = 120 N
Force to the left (Fₗ) = 80 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 120 – 80
Fₙ = 40 N to the right.
From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.