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liberstina [14]
3 years ago
11

A train moves at constant velocity of 50km/h. how far will it move in 0.5h?

Physics
1 answer:
Kobotan [32]3 years ago
7 0
The formula for getting the distance will be distance = speed x time
                                                                               D = S x T
   speed or velocity = 50km/h
                       time = 0.5 h
    the equation will be done directly because it's already in it's SI units
            distance = 50km/h  x  0.5h
            hour cancels hour and the equation remains = 50km x 0.5
                               Ans = 25 km
the train will move 25 km far in 0.5h
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if you crash your car how could you decrease the damage to you or the car using the concept of impulse
kotykmax [81]

Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

3 0
3 years ago
On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?
Alenkasestr [34]

Answer:

<h2>2.35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

<h3>2.35 N</h3>

Hope this helps you

8 0
3 years ago
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
A ball is dropped off the balcony of a hotel room and it takes 2.8s to fall to the ground . how high above the ground is the bal
RideAnS [48]

The height of the ball above the ground is 38.45 m

First we will calculate the velocity of the ball when it touch the ground by using first equation of motion

v=u+gt

v=0+9.81×2.8

v=27.468 m/s

now the height of the ground can be calculated by the formula

v=√2gh

27.468=√2×9.81×h

h=38.45 m

5 0
3 years ago
PLS HELP !!
Ira Lisetskai [31]
A because it’s basic kinetic example
4 0
3 years ago
Read 2 more answers
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