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Zarrin [17]
3 years ago
15

A vehicle has an initial velocity of 30 meters per second. 30 seconds later, it is traveling at a velocity of 60 meters per seco

nd. What is the acceleration of the vehicle?
Physics
1 answer:
Over [174]3 years ago
3 0
1 m/s^2 (60-30) then divide by 30 secs :)
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A steel ball bearing is released from a height H and
ArbitrLikvidat [17]

Answer:

ELASTIC collision

kinetic energy is conservate

Explanation:

As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.

As the floor does not move the conservation of the moment

            po = pf

            -mv1 = m v2

- v1 = v2

So the speed with which it descends is equal to the speed with which it rises

Therefore the kinetic energy of the ball before and after the collision is the same

4 0
3 years ago
What is the specific heat of a material that has a mass of 5 grams and increases 125ºC when it receives 150J of heat?
rjkz [21]

Answer:

0.24 ? I hope that was the answer you were looking for.

Explanation:

6 0
3 years ago
Read 2 more answers
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration
umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

F=qvB     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

F=qvB=ma       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0
3 years ago
An elevator (mass 4125 kg) is to be designed so that the maximum acceleration is 0.0400g.
fiasKO [112]
The way to do this is very easy so do 4125 x 2 = ? then the ? will be times by 2 again after the answer to both of those is your answer!!!
8 0
3 years ago
A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
pochemuha

Answer:

2856.96 J

0

0

\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

6.78822 m/s

Explanation:

v_i = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

At height y = 0 the potential energy is 0 as

P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0
3 years ago
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