All categories

3 years ago

Calculate the force needed to move a 2kg mass with an acceleration of 5ms-2

Physics

1 answer:

kap26 [50]3 years ago

7 0

Answer:

$\Huge \boxed{\mathrm{10 \ N}}$

Explanation:

$\sf Force \ (N)=mass \ (kg) \cdot acceleration \ (ms^{-2})$

$F=ma$

The mass is 2 kg.

The acceleration is 5 ms⁻².

$F = 2 \cdot 5$

$F=10$

The force is 10 N.

You might be interested in

I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.

MakcuM [25]

Answer:

210

Explanation:

A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?

S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.

S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.

5 0

2 years ago

The odometer of a car changes from 1048 km to 1096 km in 40

Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0

2 years ago

A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

$\vec{F_B} = q(\vec{V}\times \vec{B})$

$F_B = q|v||B| sin\theta$

Here,

q = Charge

v = Velocity

B = Magnetic field

$\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}$

Our values are given as,

$\theta = 35.7\°$

$q = 22.5*10^{-9}C$

$B = 1.05*10^{-5}T$

$v = 3.11m/s$

Replacing,

$F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)$

$F_B = 4.224*10^{-13}N$

Therefore the size of the magnetic force acting on the bumble bee is $4.22*10^{-13}N$

3 0

3 years ago

A feather, a marble, and a cannonball are dropped from the same height at

Vika [28.1K]

The correct answer is that they would all hit the ground at the same time. If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube.

7 0

3 years ago

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m$

The displacement of the ball on the moon is 7.04 m

6 0

3 years ago