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2 years ago

A dog runs after a car. The car is travelling at an average speed of 5 m/s.

Physics

2 answers:

n200080 [17]2 years ago

3 0

Answer:

<h2>The answer is 4 s</h2>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

where

d is the distance

v is the velocity

From the question we have

$t = \frac{20}{5} \\$

We have the final answer as

Hope this helps you

Lina20 [59]2 years ago

3 0

Answer:

4 secs

Explanation:

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How can I solve the following statement?

Firlakuza [10]

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d = 9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

$E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C$ leftwards

The electric field due to the charge q' at midpoint is

$E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C$

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

7 0

2 years ago

What is the momentum of a 0.5 kg ball which is thrown at 20 m/s?

tatuchka [14]

A =-16.7 Ns B= -167.N

7 0

2 years ago

Your friend clams that objects do not have to be thouching for a magnetic force to cause motion. How would you support your frie

Leni [432]

Yes because something that has been electrically charged can make other things move without touching them ( this is called force without contact)
Hoped this helped :)

4 0

2 years ago

At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

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3 years ago

my choice1A child pushes a toy truck up an inclined plane. The lifting force is...Select one:ut ofO a. the weight of the truck.O

mash [69]

When pushing a toy truck up an inclined plane, the force that makes the truck goes up is the force that the child uses to push the truck. (The force that pushes

Therefore the correct option is C.

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7 months ago