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4 years ago

What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2

Physics

1 answer:

Scorpion4ik [409]4 years ago

4 0

Answer:

20 N

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma

= 5 × 4 = 20 N

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3 years ago

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Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the

Maksim231197 [3]

Answer:

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

Hence the acceleration due to gravity on the planet is 1.40625m/s²

6 0

3 years ago

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

8 0

3 years ago

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

3 years ago

What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2​

What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2