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2 years ago

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A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s. The frictional force between the bl

ock and the floor is 10 N. The net work done on the block in 1 minutes is most nearly:

1 answer:

horrorfan [7]2 years ago

7 0

Answer:

5× 3 over 10 = 1.5

Explanation:

F = 1m/s × kg over n

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If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level loc

geniusboy [140]

Complete question:

The classic Goodyear blimp is essentially a helium balloon— a big one, containing 5700 m³ of helium. If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20°C.

Answer:

52.4 kN

Explanation:

The helium at 20°C has a density of 0.183 kg/m³, and the cargo load is the weight of the system, which consists of the envelope, the gondola, and the helium.

The helium mass is the volume multiplied by the density, thus:

mHe = 5700 * 0.183 = 1043.1 kg

The total mass is then 5343.1 kg. The weight is the mass multiplied by the gravity acceleration (9.8 m/s²), so:

W = 5343.1*9.8

W = 53362.38 N

W = 52.4 kN

5 0

2 years ago

Read 2 more answers

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

6 0

2 years ago

Which diagram shows how Rachel can see a candle flame?

SIZIF [17.4K]

Answer:

C

Explanation:

If the arrows represent light rays, then Rachel sees a candle flame when the light released by the flame is received by her eyes.

5 0

2 years ago

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

Gelneren [198K]

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

$mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s$

So, the final velocity of the blue ball is 1.21 m/s.

4 0

2 years ago

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders,

S_A_V [24]

Answer:

The total number of small cylinder = 7.

Explanation:

Lets take

Radius of the large cylinder = R

length = L

L = 10 R

The total area A = 2 π R² + π R L

The length of the small cylinder = l

The number of small cylinder = n

L = n l

The total area of small cylinders

A'=n (2 π R² + π R l)

As we know that emissive power given as

P = A ε σ T⁴

For large cylinder

P = A ε σ T⁴ -----------1

For small cylinders

P'=A' ε σ T⁴ ------2

From 1 and 2

Given that

P'= 2 P

A' ε σ T⁴ =2 A ε σ T⁴

A'=2 A (All others are constant)

n (2 π R² + π R l) =(2 2 π R² + π R L)

n (2 R² + R l) = (2 R² + R L)

$n(2R^2+R\times \dfrac{L}{n}) = 2(2R^2+RL)$

L = 10 R

$n(2R^2+R\times \dfrac{10R}{n}) =2 (2R^2+R\times 10R)$

$n(2+\dfrac{10}{n}) =2( 2+ 10)$

2 n +10 = 2 x 12

2 n +10 = 24

2 n = 24 -10

2 n = 14

n = 7

The total number of small cylinder = 7.

3 0

2 years ago