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ollegr [7]
3 years ago
12

two toy cars can roll without friction on a horizontal surface. Car A is a 5 kg. Car B is 0.5 kg. Both cars start from rest. Equ

al sized forces are used to push each cart forward for a time of 1 second. After 1 second the force is removed. What is the difference in magnitude of the momenta of the two cars, right after the force is removed?Choices:A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)B. The momentum of car A (5kg) is LESS than that car B (0.5)C. The momentum of car A (5kg) is MORE than that car B (0.5)D. 5.5 kg m/s
Physics
1 answer:
Llana [10]3 years ago
8 0

Answer:

A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)

Explanation:

The moment, or impulse formula of the same forces acting on both car within 1 second is

\Delta p = F * \Delta t

In our case the forces are the same, the time duration of force acting on the cars are the same. Therefore, their momentum right after the force must also be the same.

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A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the
kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}

Now for time:

\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}

Solving for t,

\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}

The total time the kangaroo takes in the air is 2.3s.

3 0
1 year ago
A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 22.0 N to keep this 7.00 kg
xxMikexx [17]

Answer:

B . 68.7 N

Explanation:

Given in the question that;

Force = 22 N

Mass = 7.0 kg

Velocity = 4.0 m/s

Gravitational force is the weight of wooden block , calculated as mass times acceleration due to gravity.

F= m* a

F= 7 * 9.81

F= 68.67 N

F= 68.7 N

7 0
3 years ago
Which diagram models the position of a soccer ball when it has the greatest amount of gravitational potential energy?
posledela

Answer:

C

Explanation:

Diagram C is the correct answer, because the ball is at the point with the highest height relative to the ground, in this way all the kinetic energy has been transformed into potential energy.

We must remember that potential energy is defined as the product of mass by gravity by height

Ep = m*g*h

where:

m = mass [kg]

g = gravity acceleration [m/s²]

h = elevation [m]

So when we have a great value for h in the above equation, we will have a big value for potential energy.

8 0
3 years ago
Help ~~~ science question
lakkis [162]
Where is the message
6 0
3 years ago
If your car was traveling at 60 mph, that is about 27m/s. How long would it take a car traveling at 27m/s to slow to a stop, if
RUDIKE [14]

Answer:

vf=at+v0. vf=0

-v0=at

t=-v0/a=(-27m/s)/(-8m/s^2)=3.4s

6 0
3 years ago
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