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kupik [55]
3 years ago
7

PLS HELP ITS URGENT HELP PLSSS

Physics
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

heyyy will u be my friend

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How do force affect the acceleration of a body?<br><br>​
Arada [10]

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

3 0
2 years ago
The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​
Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0
2 years ago
If a proton were released in the electric field above, what direction would it move?
Afina-wow [57]

Answer:

d because the proton would move towards the negative plate

Explanation:

5 0
3 years ago
Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plate
Daniel [21]

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :

$E=\frac{V}{D}$

Differentiating on both the sides with respect to time, we get

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

      $=\frac{1}{1.4 \times 10^{-3}} \times 110$

      $=7.85 \times 10^4$  V/m-s

8 0
3 years ago
3.00 m^3 of water is at 20.0°C.
romanna [79]

Answer:

\triangle V = 0.02484m^3

Explanation:

Given

V_1 = 3.00m^3 --- initial volume

T_1 = 20.0^oC --- initial temperature

T_2 = 60.0^oC --- final temperature

\gamma = 207*10^{-6 ---  coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

\triangle V = \gamma * V_2 * (T_2 - T_1)

So, we have:

\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)

\triangle V = 207 * 10^{-6} * 3.00 * 40.0

\triangle V = 0.02484m^3

The volume will expand by 0.02484m^3

7 0
2 years ago
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