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3 years ago

5

- 3000 J of work is done on a 2kg object over a distance of 10m. What is the force acting on the object?

1 answer:

SpyIntel [72]3 years ago

7 0

The answer is B. 300N

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Which of the following is not an application of Doppler technology?

Jlenok [28]

The correct answer to the question above is the third option; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology. If the Doppler technology is being used in medical field, it would be for the ultrasound of the heart and blood vessels for examination.

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3 years ago

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Why shadow is black in colour ?

lions [1.4K]

Answer:

Absence of light causes shadow

Explanation:

The shadow is of the black colour because it is formed by the white light from the sun. Black is the opposite colour of white. In case you use different colour of light to cast a shadow you will get a shadow of opposite colour of light used to cast shadow.

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3 years ago

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The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

7 0

3 years ago

A soccer player icks a rock horizontally off a 40m high cliff into a pool f water if the player hears the sound of the splash s

Semenov [28]

Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 40}{9.8}}$

t = 2.857 s

distance

d = v x t

d = v x 2.875

time traveled by the sound before reaching the player

$t_0 = t - t_{fall}$

$t_0 = 3 - 2.875$

$t_0 = 0.143 s$

distance traveled by the wave in this time'

r = 0.143 x 343

r= 49.05 m

now,

we know.

d² + h² = r²

d² + 40² = 49.05²

d =28.387 m

v x 2.875=28.387 m

v = 9.936 m/s

7 0

3 years ago

The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks

zavuch27 [327]

Answer:

After 2.0s the angular momentum is $L= 2(4A+3B+2C+D)x$

Explanation:

Let us call forces acting on the rod, A, B, C, and D, and the separation between them $x$ .

Then, the torque due to force A is

$\tau_a = 4Ax$ ,

due to the force B

$\tau_b = 3Bx$ ,

due to force C

$\tau_c = 2Cx$ ,

and the torque due to force D is

$\tau_d = Dx$ .

Therefore, the total torque on the the stick is

$\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d$

$\tau_{tot} =4Ax+3Bx+2Cx+Dx$

$\tau_{tot} =x(4A+3B+2C+D)$

Now, this torque causes angular acceleration $\alpha$ according to the equation

$I \alpha = \tau_{tot}$

where $I$ is moment of inertia of the stick and it has the value

$I = \dfrac{1}{3} m(4x)^2$

Therefore the angular acceleration is

$\alpha = \dfrac{\tau_{tot} }{I}$

$\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }$

$\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}$

Now, the angular momentum $L$ of the stick is

$L = I\omega$ ,

where $\omega$ is the angular velocity.

Since $\omega = \alpha t$ , we have

$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

$L= (4A+3B+2C+D)x t$

Therefore, $t = 2.0s$ , the angular momentum is

$\boxed{ L= 2(4A+3B+2C+D)x. }$

5 0

3 years ago