What is the HORIZONTAL component of a vector with a magnitude of 125 m/s and an angle of 25 degrees?

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

madreJ [45]

Hello!

a) What is the change in momentum?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

$\boxed{a=\frac{V-Vi}{t} }$

Like you see, the final velocity will be zero, so:

$a = \dfrac{0m/s-40m/s}{4s}$

$a = \dfrac{-40m/s}{4s}$

$a = -10\ m/s^{2}$

Like you see, the aceleration applicate by superman is of -10 m/s^2.

If the aceleration is negattive, means the velocity will decrease.

b.) What is the magnitud of the force wich superman exerts on the train?

For calculate the force applicate for superman, lets applicate the second law of Newton:

$\boxed{F=ma}$

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of -80000 Newtons.

When the force is negative means the force applicated in the another direction of the object what is traveling.

8 0

3 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

now here we can say

$m_1 = 10 g$

$v_{1i} = 0$

$m_2 = 550 g$

$v_{2i} = 3.5 m/s$

now here we can say

$10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}$

$192.5 = v_{1f} + 55 v_{2f}$

now by coefficient of restitution

for elastic collision we know that e = 1

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

$v_{2f} - v_{1f} = 0 - 3.5$

now by solving the two equation

$56v_{2f} = 189$

$v_{2f} = 3.375 m/s$

also we know that

$v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s$

so final speed of the nail is 6.875 m/s

6 0

3 years ago

Read 2 more answers

2) In coming to a stop, a car leaves skid marks on a road that are 40 m long.

Effectus [21]

Lolilolololilolollololililili

5 0

3 years ago

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi

Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0

3 years ago

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = $\frac{1}{2}mu^2$

Final kinetic energy = $\frac{1}{2}mv^2$

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2$

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2$

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0

4 years ago