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3 years ago

9

what is the work done by gravity when a 2.0kg ball falls to the floor from the height of 1.50m? Is it positive or negative? Expl

ain

1 answer:

Ratling [72]3 years ago

5 0

Positive because the ball is falling downwards.
Work done(Wd)= force x distance travelled in the direction of the force,
Wd=(2x9.8)x1.5

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Dynamics Newton's 1st & 2nd Laws

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Ummm what does this even mean I need a picture to help u

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2 years ago

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

<em>Mass of the child, m = 16 kg</em>
<em>Length of the incline, L = 2 m</em>
<em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

$sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m$

The gravitational potential energy of the child at the top of the incline is calculated as;

$P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J$

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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1 year ago

A spaceship accelerates from 0m/s to 40m/s in 5 seconds. What is the acceleration of the spaceship

nadya68 [22]

Acceleration = change in velocity/time
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3 years ago

A construction worker puts 20J of energy in to one strike of his hammer on the head of a nail. The energy transferred to driving

Slav-nsk [51]

Answer:

efficiancy=40 percent

Explanation:

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2 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

2 years ago