Answer:
[tex]V=14m/s[/tex]
Explanation:
From the Question we are told that
Mass of A and B is 60kg
Speed of A=2m/s
Speed of B=1m/s
Mass of bag =5kg
Generally the momentum of the astronaut A and bag is mathematically given as
[tex]M_A=(60+5)*2[/tex]
[tex]M_A=130kgm/s[/tex]
Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag
Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1
Therefore
[tex]130=(60*1)=(5*v)[/tex]
[tex]V=14m/s[/tex]
To avoid collision with B, A throws the bag with a speed of 26 m/sec.
Given-
The mass of the two astronauts A and B is 60 kg.
The mass of astronaut A is 65 kg with the box.
The speed of astronaut A is 2 m/sec.
The speed of astronaut B is 1 m/sec.
The mass of the bag is 5 kg.
The momentum of a body is the product of its mass and velocity. Thus the momentum of astronaut A is,
[tex]M_{A} =m_{a} \times v_{a}[/tex]
[tex]M_{A} =(60+5)\times2[/tex]
[tex]M_{A} =130[/tex]
To avoid the collision with B, A throws the bag with a speed v. For this the momentum after the trough must be equal to the momentum of A. Therefore,
[tex]M_{A} =5\times v_{r}[/tex]
[tex]130 =5\times v_{r}[/tex]
[tex]v_{r}= \dfrac{130}{5}[/tex]
[tex]v_{r}=26[/tex]
Hence, to avoid collision with B, A throws the bag with a speed of 26 m/sec.
For more about the velocity, find the link below-
https://brainly.com/question/862972
Two similarities between balanced and unbalanced forces.
Explanation:
For balanced forces, the severity of the 2 factors is equal, while the magnitude of the 2 factors is unequal in the situation of unbalanced forces. The three separate forces operate in opposing ways in balanced forces. In unbalanced forces, on the other hand, independent forces either behave in the same or reverse direction.
What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth
Answer:
c
Explanation:
what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a distance between Dam of 0.002 M
Answer:
[tex]from \: newton \: law \: of \: gravitation \\ F = \frac{GMm}{ {r}^{2} } \\ G = 6.67 \times {10}^{ - 11} \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F = \frac{(6.67 \times {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} } \\ = 2.75 \times {10}^{ - 6} N[/tex]
A north magnetic pole brought near a south magnetic pole will...
A)
cancel the south pole.
B)
attract the south pole.
C)
repel the south pole.
At the surface of a certain planet, the gravitational acceleration g has a magnitude of 20.0 m/s^2. A 22.0-kg brass ball is transported to this planet.
What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?
Answer:
a, 22 kg and 22 kg
b, 215.8 N and 440 N
Explanation:
a
The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg
b
The weight of any object factors in the acceleration due to gravity of the said area(or planet).
W = mg, with m being the mass and g being the acceleration due to gravity.
On earth
W = 22 * 9.81 = 215.8 N
On the said planet,
W = 22 * 20 = 440 N
Do therefore, the weight is 215.8 N on earth and 440 N on the planet
pls help, i have no idea where to start
Answer:
Use equation 1/2 * m * v²
Explanation:
where m = mass in kg
and v = velocity in m/s
plug accordingly.
The x-axis of a position-time graph represents
Answer:
The y-axis represents position relative to the starting point, and the x-axis represents time.
Explanation:
Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.
Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]
what effects do coil of wire on a compass after it has been connected to a battery
Explanation:
The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.
Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she arrives at the mark in 20min?
A box moves 5\,\text m5m5, start text, m, end text horizontally when a force, F =10\,\text NF=10NF, equals, 10, start text, N, end text is applied at an angle, \theta =150\degreeθ=150°theta, equals, 150, degree as shown below. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. What is the work done on the box by FFF during the displacement?
Answer:
??
Explanation:
not completely sure hold on
Answer:
-43J
Explanation:
F eeeeeeeereeeeeee points
Answer:
Yayyyyyy thank you ≥﹏≤.!!!!!!!!!!!!!
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B moves forward at 10 m/s and car A is at rest. What fraction of the initial kinetic energy is lost in the collision?
Given :
Initial speed of car A is 15 m/s and initial speed of car B is zero.
Final speed of car A is zero and final speed of car B is 10 m/s.
To Find :
What fraction of the initial kinetic energy is lost in the collision.
Solution :
Initial kinetic energy is :
[tex]K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}[/tex]
Final kinetic energy is :
[tex]K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}[/tex]
Now, fraction of initial kinetic energy loss is :
[tex]Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25[/tex]
Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .
Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?
Mid-fielder
Center
Forward
Guard
Answer:
Center
Explanation:
The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.
Answer:
C
Explanation:
1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
Learn more about the work done here:
https://brainly.com/question/13662169
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.
Answer:
The unknown force will be 18.116 lb.
Explanation:
Given that,
Three forces act on a flange as shown in figure.
The net force acting on the flange is a minimum.
[tex]\dfrac{dF_{net}}{df}=0[/tex]
We need to calculate the unknown force
Using formula of net force
[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]
Put the value into the formula
[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]
[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]
The magnitude of net force,
[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]
[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]
[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]
[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]
On differentiating w.r.to F
[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]
[tex]0=2F+36.232[/tex]
[tex]F=-\dfrac{36.232}{2}[/tex]
[tex]F=-18.116\ lb[/tex]
Negative sign shows the direction of force which is downward.
Hence, The unknown force will be 18.116 lb.
Following are the solution to the given question:
Calculating the Net force on flange:
[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]
Calculating the magnitude:
[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]
Differentiate the value
[tex]\to \frac{F_{net}}{df}=0 [/tex]
[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]
Learn more:
brainly.com/question/15323099
Fill in the blanks in the following paragraph to correctly identify the properties of devices that use electromagnetism. (1 point)
A(n) ____________________ changes mechanical energy into ____________________. A(n) ____________________ changes electric energy into ____________________. A(n) ____________________ changes the voltage of an alternating current. A(n) ____________________ transformer has more loops in the primary coil than in the secondary coil. A(n) ____________________ transformer has ____________________ loops in the primary coil than in the secondary coil.
Answer:
this answer you question
Answer:
generatorelectricalmotormechanicaltransformerstep-downstep-upfewerExplanation:
A(n) generator changes mechanical energy into electrical. A(n) motor changes electric energy into mechanical. A(n) transformer changes the voltage of an alternating current. A(n) step-down transformer has more loops in the primary coil than in the secondary coil. A(n) step-up transformer has fewer loops in the primary coil than in the secondary coil.
Hope it's helpful!
Can yall help me please
Answer:
physical
Explanation:
it is a real life model that was built
6. How do the data support your claim above? Explain your statement above. Be sure to refer to specific pieces of data from your experiment that support your argument.
Answer:
hope this will help
Explanation:
if your title doesn't have enough information, make a list of the key words ... If you are not sure what should be included in each summary sentence, use ... Often you can present the hypothesis and the supporting reasoning in one paragraph. ... Describe specific parts of the procedure or data that contributed to your learning.
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/191502-7how-do-the-data-support-your-claim-above-explain-your-statement.html
What is the net force?
60N left
4ON, right
60N, right
0N
Answer:
60n left
Explanation:
Do
Help me
Shown is a picture of two toy trains getting ready to collide. If the collision is non-elastic, what is the total momentum of
the train cars after the collision?
a
-10 kg·m/s
b
10 kg·m/s
c
70 kg·m/s
d
50 kg·m/s
Answer:
A: -10 kg•m/s
Explanation:
6•5= 30
-2•20=-40
-40+30= -10
-10 kg•m/s
A 20N cart is being pulled to the right with an applied force of 20N at an angle of 40°. The normal force on the cart is 7N and the friction is 3N. Draw the free body diagram. What is the net force in the x and y direction of the cart?
Answer:
F_{net-x} = 12.32 N , F_{net-x} = -0.14 N
Explanation:
For this case we apply Newton's second law on each axis, see attachment
X axis
F cos 40 - fr = m a
Y Axis
F sin 40 +N - W = 0
the net force is
X axis
[tex]F_{net-x}[/tex] = F - fr
F_{net-x} = 20 cos 40 - 3
F_{net-x} = 12.32 N
Y Axis
F_{net-y} = F sin 40 + N - W
F_{net-x} = 20 sin 40 + 7 - 20
F_{net-x} = -0.14 N
we see that the force in this axis is approximately zero
The density of a solid or liquid material divided by the density of water is called
Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
The correct answer is 40 or - 40??
Answer:
40 cm
hope it's help you but you can choose if your ans. is -40 cm
I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST
Answer:
u will need 1/2 cells
Explanation:
What height would a 4 kg book need to be to have a potential energy of
235.2 J on earth?*
need help!!
Answer:
5.99 m = 6 m
Explanation:
PE = m*g*h
235.2 J = (4 kg)(9.81 m/s^2)(h)
h = (235.2 J)/(9.81*4)
h = 5.99 m
h = 6 m
What is the force of gravitational attraction between an object with a mass of 2kg and another object that has a mass of 2kg and a distance between them of 2.5m
Answer:
1.7x10^-10 N
Explanation:
F = G [(m_1)(m_2)]/(r^2)
F = force
G = Gravitational constant 6.67433x10^-11 (N*m^2)/kg^2
m_1 = mass first object
m_2 = mass second object
r = radius between the 2 objects
F = G[(2 kg*2 kg)/(1.25 m)^2]
F = 1.7x10^-10 N
What does the Big Bang theory explain?
A. How time began
B. How the universe began
C. Both A and B
D. Neither A nor B
Answer:
how the universe began
Explanation:
time never stops or starts, we only think it starts when we start recording it
how does a switch work in a circuit
An electric switch is a device that interrupts the electron flow in a circuit. Switches are primarily binary devices: either fully on or off and light switches have a simple design. When the switch is turned off, the circuit breaks and the power flow is interrupted. Circuits consist of a source of power and load.
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.