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3 years ago

A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?

Physics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

$p=m.v$

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

$\displaystyle v=\frac{p}{m}$

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

$\displaystyle v=\frac{6}{0.15}=40\ m/s$

The velocity is directed to the south

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A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler

Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

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3 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with water to the rim. (a) Determine

pashok25 [27]

Answer:

44100 N

Explanation:

Each wall will have dimension of 4 m x 1.5 m

Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m

pressure at CM = h d g , h = .75 , d ( density of water = 10³ )

pressure at CM = .75 x 10³ x 9.8

= 7350 N / m²

Total force on each wall

= pressure x area

= 7350 x 4 x 1.5

= 44100 N Ans

b ) If h = 1.5 x 2 = 3

Pressure = hdg

1.5 x 10³ x 9.8

= 14700 N / m²

Force

= pressure x area

14700 x 3 x 4

= 176400 N

Which is 4 times 44100 N

So force will quadruple.

It is so because both area and height have become twice.

3 0

3 years ago

"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s

Tamiku [17]

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

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3 years ago