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3 years ago

Calculate the speed of an object which Travels 30 km distance in 4 hours

Physics

1 answer:

nexus9112 [7]3 years ago

7 0

Answer:

The object is traveling 7.5 km per hour.

Explanation:

30/4 = 7.5

To check do 7.5 * 4 and you get 30

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A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

DochEvi [55]

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

Initial velocity of the cart = 0.2 m/sec

And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time $\Delta t=0.1sec$

Change in momentum is given by

$\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec$

Now force is given by

$F=\frac{\Delta P}{\Delta t}=\frac{0.4875}{0.1}=4.875N$

Net force will be 4.875 N

6 0

3 years ago

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m

4 0

3 years ago

Read 2 more answers

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago

The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c

Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0

3 years ago

The force of gravity is twice as great on a 2-kg rock as on a 1-kg rock. Why then, dose the 2-kg rock not fall with twice the ac

Phoenix [80]

Answer:

because all objects fall at a rate of 9.8m/s²

8 0

3 years ago

Calculate the speed of an object which Travels 30 km distance in 4 hours​

Calculate the speed of an object which Travels 30 km distance in 4 hours