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nekit [7.7K]
3 years ago
7

HI PLEASE I REALLY NEED HELP....

Physics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

(B) 5.34 m/s

Explanation:

We can solve this with forces or with energy.

First, let's solve using forces.

The sum of the forces on the sphere in the y direction is:

∑F = ma

T − Mg = M(-a)

T = Mg − Ma

The sum of the forces on the block parallel to the incline is:

∑F = ma

T − F − mg sin θ = ma

Substituting and solving for a:

Mg − Ma − F − mg sin θ = ma

Mg − F − mg sin θ = (m + M) a

a = (Mg − F − mg sin θ) / (m + M)

a = (4×9.8 − 2 − 3×9.8×⅗) / (3 + 4)

a ≈ 2.79 m/s²

So the velocity after 5 meters of travel is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(2.79)(5)

v ≈ 5.29 m/s

Using energy:

Potential energy = Potential energy + Kinetic energy + Work

MgH = mgh + ½ Mv² + ½ mv² + Fd

(4)(9.8)(5) = (3)(9.8)(3) + ½ (4)v² + ½ (3)v² + (2)(5)

196 = 88.2 + 3.5 v² + 10

v ≈ 5.29 m/s

The closest answer is B, 5.34 m/s.

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Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box
mariarad [96]

Answer:

Bill's motor

Explanation:

Bill's motor lifts a box 0.39 metres in 2 seconds.

Nageen's motor lifts the same box 0.35 metres in 2 seconds.

Power = Work done/time

or

P=\dfrac{mg\times h}{t}

Power of bill's motor,

P_1=\dfrac{m\times 9.8\times 0.39}{2}\\\\P_1=1.91m\ Watts

Power of Nageen's motor,

P_2=\dfrac{m\times 9.8\times 0.35}{2}\\\\P_2=1.71m\ Watts

So, Bill's motor applied more power to the box.

5 0
3 years ago
If you can answer both, please do. But if you can't, just answer one.
Setler [38]

Answer:

1.The force required to stop the shopping cart is, F = 12.25 N

Explanation:

Given data,

The mass of the shopping cart, m = 7 kg

The initial velocity of the shopping cart, u = 3.5 m/s

The final velocity of the shopping cart, v = 0 m/s

The time period of acceleration, t = 2 s

The change in momentum of the cart,

                                      p = m(u - v)

                                         = 7 (3.5 - 0)

                                         = 24.5 kg m/s

The force is defined as the rate of change of momentum. To stop the shopping cart, the force required is given by the formula

                                           F = p / t

                                               = 24.5 / 2

                                               = 12.25 N

Hence, the force required to stop the shopping cart is, F = 12.25 N

2.

We have: F = m × v/t

Here, m = 8500 Kg

v = 20 m/s

t = 10 s

Substitute their values into the expression,  

F = 8500 × 20/10

F = 8500 × 2

F = 17000 N

In short, final answer would be 17000 N

Hope this helps!!

7 0
3 years ago
The acceleration due to gravity on Jupiter is 23.1 m/s2, which is about twice the acceleration due to gravity on Neptune.
polet [3.4K]

Answer:

I believe its C

Explanation:

3 0
3 years ago
Read 2 more answers
How is Newton’s first law demonstrated by your zip line
Nikitich [7]

Newton's First Law: An object in motion stays in motion, an object at rest stays at rest unless an unbalanced force acts upon it. For example, when zipping down the zip line you will stay in motion unless an outside force interferes. ... The more mass the more force needed.

4 0
3 years ago
when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of
nata0808 [166]

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

E = \omega NBA |sin \omega t|

Here,

\omega = Angular frequency

N = Number of turns

B = Magnetic field

The emf has its maximum value when sin \omega t = \pm 1

Thus the amplitude of the emf is

E = \omega NBA

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

E \propto \omega

Further it can be written as follows,

\frac{E_1}{E_2} \propto \frac{\omega_1}{\omega_2}

E_2 = \frac{\omega_2}{\omega_1}E_1

E_2 = \frac{10rad/s}{15rad/s}(27.0V)

E_2 = 18V

Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V

5 0
3 years ago
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