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Charra [1.4K]
3 years ago
10

A motorcycle has a mass of 2.50×10^2. A constant force is exerted on it for 60.0s. The motorcycles initial velocity is 6.00 m/s

and its velocity 28.0 m/s. What is it’s change in momentum?
Physics
1 answer:
Margarita [4]3 years ago
8 0

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have:

m=2.50\cdot 10^2 kg (mass of the motorcycle)

v=28.0 m/s (final velocity)

u=6.0 m/s (initial velocity)

Therefore, the change in momentum is

\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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Dimension of radius of sphere​
Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = \frac{4}{3}\pir^{3}

In this equation \pi is a dimensionless quantity, and the unit of v is m^{3}.

But, metre is a measure of length, thus it has a dimension of L.

So that,

m^{3} ≅ L^{3}

Then,

L^{3} = r^{3}

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0
3 years ago
A baseball rolls off of a .7 m high desk and strikes the floor .25 m always how fast was the ball rolling
labwork [276]

Answer:

the ball's velocity was approximately 0.66 m/s

Explanation:

Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.

Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.

Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:

0.7 = (1/2) g t^2

solve for t:

t^2 = 1.4 / g

t = 0.3779  sec

which we can round to about 0.38 seconds

No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:

horizontal distance covered = vi * t

0.25 = vi * (0.38)

solve for vi:

vi = 0.25/0.38  m/s

vi = 0.65798  m/s

Then the ball's velocity was approximately 0.66 m/s

4 0
3 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
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Please help me on questions 3 and 6, thank you! :D I'll give brainliest!
Anarel [89]

Answer:

3. if you increase your mass you also increase the gravitational pull

6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.

Explanation:

i hope this helps-

4 0
3 years ago
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Which statement correctly stated kepler's 3rd law of planetary motion?
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When you square the "year" of each planet and divide it by the cube of its distance, or axis from the sun, the number would be the same for all the planets
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3 years ago
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