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Mariana [72]
2 years ago
8

An illustration of a ball sitting at the top of a hill of height labeled h Subscript 1 Baseline = 2 m. A the the bottom of the h

ill it levels off and the leveled surface is at a height labeled h Subscript 2 Baseline = .5 m. A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10 m/s2 and round the answer to the nearest tenth. m/s
Physics
2 answers:
lesya [120]2 years ago
7 0

Answer:

5.5 m/s

Explanation:

Right on Edge

IRISSAK [1]2 years ago
5 0

Answer:

5.5

Explanation:

edge 2021

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Most of the substances around you are _______
monitta

Answer:

gravity

Explanation:

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Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They
timama [110]

Answer:

c. 48 cm/s/s

Explanation:

Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?

from newtons second law of motion ,

which states that change in momentum is directly proportional to the force applied.

we can say that

f=m(v-u)/t

a=acceleration

t=time

v=final velocity

u=initial velocity

since a=(v-u)/t

f=m*a

force applied is F

m =mass of the object involved

a is the acceleration of the object involved

f=m*48.........................1

in the second case ;a mass of 2M when acted upon by a net force of 2F

f=ma

a=2F/2M

substituting equation 1

a=2(M*48)/2M

a=. 48 cm/s/s

6 0
3 years ago
Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
Iteru [2.4K]

Explanation:

Let m_p is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

\dfrac{mv^2}{r}=qvB

r is the radius of path,

r=\dfrac{mv}{qB}

Time period is given by :

T=\dfrac{2\pi r}{v}

T=\dfrac{2\pi m_p}{qB}

Frequency of proton is given by :

f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}

The wavelength of radiation is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{2\pi m_pc}{qB}

So, the wavelength of radiation produced by a proton is \dfrac{2\pi m_pc}{qB}. Hence, this is the required solution.

3 0
3 years ago
A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi
inn [45]

Answer:

\Phi_{E} = E\pi r^2 \omega t

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

\Phi_{E} = \vec{E}\vec{A}

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

\Phi_{E} = EA\cos(\theta)

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

\theta = \omega t

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

\Phi_{E} = E\pi r^2 \omega t

3 0
3 years ago
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