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3 years ago

Samiyah pushes a cart with 20 newtons of force. The cart weighs 10 kilograms. How quickly does the cart accelerate?

Physics

2 answers:

love history [14]3 years ago

3 0

Answer:

10 newtons

Explanation: 10 of the 20 cancels out and the other 10 is force.

Nadusha1986 [10]3 years ago

3 0

Answer: a = 2 m/s²

Explanation: Acceleration is equal to force over mass, a = F/m

a= 20N / 10 kg

= 2 m/s²

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3 years ago

Read 2 more answers

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

3 years ago

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

Artist 52 [7]

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s ................1

we know acceleration will be

a = $\frac{-u^2}{2\times S}$

Then we have

Force = mass × acceleration .................2

m × $\frac{-u^2}{2\times S}$ = -μk mg

The puck decelerates due to the kinetic frictional force μk mg

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

4 0

3 years ago

You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction),

Rudiy27

Answer:

a) α = 0.338 rad / s² b) θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

τ = I α

fr r = I α

Now we write the translational Newton equation in the radial direction

N- F = 0

N = F

The friction force equation is

fr = μ N

fr = μ F

The moment of inertia of a saying is

I = ½ m r²

Let's replace in the torque equation

(μ F) r = (½ m r²) α

α = 2 μ F / (m r)

α = 2 0.2 24 / (86 0.33)

α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

w² = w₀² - 2 α θ

0 = w₀² - 2 α θ

θ = w₀² / 2 α

Let's reduce the angular velocity

w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s

θ = 9.634 2 / (2 0.338)

θ = 137.3 rad

Let's reduce radians to revolutions

θ = 137.3 rad (1 rev / 2π rad)

θ = 21.9 rev

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3 years ago