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3 years ago

8

If you apply a force of 10 N to a table and push it 5 m in 10 s, how much power did you deliver? Multiple choice question. A) 50

W B) 25 W C) 500 W D) 5 W

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

D) 5 W

Explanation:

Force=10N

Distance =5m

Time=10s

Power =?

But Power= Work done/Time taken

P= W/s

W= Fxd=10x5=50

P=50/10

P=5W

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

Ksivusya [100]

Answer:

The angular speed of rotation is 1.34 rad/s.

Explanation:

Given that,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We need to calculate the angular speed of rotation

Using balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Put the value of T

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Put the value into the formula

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Hence, The angular speed of rotation is 1.34 rad/s.

7 0

3 years ago

The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will

inessss [21]

Answer: weight on Jupiter = 869.75 N

mass on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁ = 347,9 N

On the Jupiter,

G₂ = mg₂

mass on the Earth = mass on the Jupiter !

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N = 869,75 N

5 0

4 years ago

53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be

marusya05 [52]

Answer:

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

here we have

$M = 4M_e$

$R = 2R_e$

we know that for earth we have

$g = 9.81 = \frac{GM_e}{R_e^2}$

now if the radius and mass is given as above

$g' = \frac{G(4M_e)}{(2R_e)^2}$

$g' = \frac{GM_e}{R_e^2}$

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

6 0

3 years ago

Can y’all please help me with this 3 part question?

klio [65]

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

$v_{f} =v_{o} +g*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

$v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)$

44100 = 20*y

y = 2205 [m]

5 0

3 years ago

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0

3 years ago