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2 years ago

Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?

Physics

1 answer:

Sliva [168]2 years ago

7 0

Answer:

$\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star$ ...

$\huge\boxed{\fcolorbox{white}{blue}{mgh=4000}}$

I hope this helps you......

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g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn

krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

T² = ( $\frac{4\pi }{G M_s}$ ) r³ (1)

in this case the period of the season is

T₁ = 93 min (60 s / 1 min) = 5580 s

r₁ = 410 + 6370 = 6780 km

r₁ = 6.780 10⁶ m

for the satellite

T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

T² = K r³

K = T₁²/r₁³

K = $\frac{ 5580^2}{ (6.780 10^6)^2}$

K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

r³ = T² / K

r³ = 86400² / 9.99 10⁻¹⁴

r = ∛(7.4724 10²²)

r = 4.21 10⁷ m

this distance is from the center of the earth

7 0

2 years ago

A student bikes to school by traveling first dN = 1.00 miles north, then dW = 0.600 miles west, and finally dS = 0.200 miles sou

kompoz [17]

Refer to the diagram shown below.

Define unit vectors along the x and y axes as respectively $\hat{i} \, and \, \hat{j}.$

Then the three successive displacements, written in component form, are respectively
$\vec{dN} = 1.0 \, \hat{j} \\
\vec{dW} = -0.6 \, \hat{i} \\ \vec{dS} = -0.2 \, \hat{j}$

The total displacement for the first leg of the trip is
$\vec{d} = \vec{dN} + \vec{dW} + \vec{dS} \\ \vec{d}= 1.0\hat{j}-0.6\hat{i}-0.2\hat{j} \\ \vec{d}=-0.6\hat{i}+0.8\hat{j}$

Answer:
$\vec{d} = -0.6\hat{i}+0.8\hat{j}$ or (-0.6, 0.8)

6 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

maria [59]

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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2 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Lynna [10]

Answer:

1) Recollapsing universe

2) critical universe

3) Coasting universe

Explanation:

According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are

1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

3) Coasting universe - in this, expansion of universe is steady and uniform

6 0

2 years ago

Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?​

Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?