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german
3 years ago
12

A 2.80 kg mass is dropped from a

Physics
2 answers:
algol [13]3 years ago
8 0
Mark Brainliest please

Answer : 41.2 J

Explanation

The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.

Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:
EP1 = EP2 + EK2
so EK2 = EP1 - EP2
and EP2 = (2.80)(9.81)(3.00) = 82.404
So EK2 = 123.606 - 82.404 = 41.2 Joules
inn [45]3 years ago
7 0

Answer:

PE = 82.32J

Explanation:

PE = m*g*h

PE = 2.80kg*9.8m/s²*3m

PE = 82.32J

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Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentu
svetoff [14.1K]

Answer:

\vec{L}=-30\frac{kgm^2}{s}\hat{k}

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

\vec{L}=\vec{r}\ X\ \vec{p}       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

\vec{L}=-30\frac{kgm^2}{s}\hat{k}

The angular momentum is -30 kgm^2/s ^k

8 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

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4 years ago
What are some ways that seed dispersal can happen
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Seeds can blow off plants in the wind and pollinate through air travel
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Goryan [66]

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Then the time it takes for the sound to travel 20 miles is

\dfrac{20\,\mathrm{mi}}{0.21\,\frac{\mathrm{mi}}{\mathrm s}}=95\,\mathrm s

3 0
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