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daser333 [38]
1 year ago
4

A 1580 Kg car is travelling at a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring

it to a halt in a distance of 50.0 m?​
Physics
1 answer:
MArishka [77]1 year ago
5 0

Answer:

Explanation:

Assuming a frictionless car on a horizontal surface.

Kinematics answer

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 15.0²) / (2(50.0)) = -2.25 m/s²

F = ma

F = 1580(2.25)

F = 3555 N

Work-Energy theorem answer

The work of the force must equal the initial kinetic energy of the car.

F(50.0) = ½(1580)15.0²

F = 3555 N

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A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?
Ratling [72]

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

3 0
2 years ago
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Select examples of simple harmonic motion that can be observed in everyday life.
liberstina [14]
A. a child gently swinging on a swing at small angles all the time
5 0
2 years ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
2 years ago
A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?
damaskus [11]
P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s
6 0
2 years ago
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You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0
2 years ago
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