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2 years ago

If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2

Physics

2 answers:

kherson [118]2 years ago

6 0

Answer:

.5 m/s2

Explanation:

Elza [17]2 years ago

4 0

Answer:

= 0.5 m/s²

Explanation:

According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore; F = ma , where F is the Force, m is the mass and a is the acceleration.

Thus; a = F/m

but; F = 5 N, and m = 10 kg

 a = 5 /10

= 0.5 m/s²

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The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired

My name is Ann [436]

Answer:

Relation between initial speed of bullet and height h is given as

$v = \frac{m + M}{m}\sqrt{2gh}$

Explanation:

As we know that system of block and bullet swings up to height h after collision

So we have

$(m + M)gh = \frac{1}{2}(m + M)v_1^2$

so we have

$v_1 = \sqrt{2gh}$

so speed of the block + bullet just after the impact is given by above equation

Now we also know that there is no force on the system of bullet + block in the direction of motion

So we can use momentum conservation

$mv = (m + M)v_1$

now we have

$v = \frac{m + M}{m}\sqrt{2gh}$

5 0

2 years ago

What is the purpose of heat index

Nookie1986 [14]

It tells how hot it really feels when the relative humidity is factored in with the actual air temperature.
hope this helps

7 0

2 years ago

Read 2 more answers

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

2 years ago

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0

3 years ago

Read 2 more answers

A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at

Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0

2 years ago