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IrinaK [193]
3 years ago
9

A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5

m?
Physics
1 answer:
Wittaler [7]3 years ago
5 0

Answer:

8.83m

Explanation:

Given parameters:

  Acceleration  = 7.8m/s²

   Initial velocity = 0m/s

  distance covered  = 5m

Unknown:

Final velocity  = ?

Solution:

The equation to solve this problem with is given as;

    V² = U² + 2aS

V is the final velocity

U is the initial velocity

a is the acceleration

S is the distance

 Input the parameters and solve;

    V² = 0² + 2 x 7.8 x 5 = 78

    V = √78 = 8.83m

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A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
What are the two ost important processes in the oxygen in and out of the atmospher
BaLLatris [955]
Decomposing - When plants and animals die, they decompose. This process uses up oxygen and releases carbon dioxide. Rusting - This is also called oxidation. When things rust they use up oxygen.
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3 0
3 years ago
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A vector starts at the point (0, 0) and ends at (3, 1). What is the magnitude and direction of the displacement?
Sindrei [870]

Answer:

tbh vector does not have any direction at all the answer is 0

6 0
3 years ago
Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona
erica [24]

Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

Mass of the person, m = 60 kg

The acceleration due to gravity on the surface of gravity is 3.7 m/s²

We need to find the weight of a person on the surface of Mercury.

Weight of an object is given by :

W = mg

So,

W = 60 kg × 3.7 m/s²

W = 222 N

Hence, the person will weigh 2222 N on the surface of Moon.

4 0
3 years ago
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