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Virty [35]
2 years ago
12

a dog runs after the car, the car is travelling at an average speed of 5 m/s, the dog runs 20 m in 5s. does she catch the car

Physics
1 answer:
ratelena [41]2 years ago
7 0

No

Explanation:

Average speed of car = 5m/s

Distance run by dog = 20m

Time taken = 5s

problem;

does the dog catches the car = ?

Solution;

let us determine speed of the dog;

   Speed = \frac{distance}{time}  = \frac{20}{5} = 4m/s

The average speed of the car is 5m/s

Average speed of the dog is 4m/s

The dog is slower than the car and it will not catch the car at this rate.

learn more:

Average speed brainly.com/question/5063905

#learnwithBrainly

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Soon after Earth was formed, heat released by the decay of radioactive elements raised the average internal temperature from 300
Oliga [24]

Answer:

97.03%

Explanation:

The equation for volumetric expansion due to thermal expansion is as follows

V/Vo=(1+γΔT)

V=final volume

Vo=initial volume

γ=coefficient of volume expansion=3.2 × 10–5 K–1

ΔT=

temperature difference

assuming that the earth is a sphere the volume is given by

V=(4/3)pi R^3

if we find the relationship between the initial and final volume we have the following

\frac{V}{Vo}  =\frac{ \frac{4}{3} \pi r^{3} }{ \frac{4}{3} \pi ro^{3}}=\frac{r^{3} }{ro^{3}}

taking into account the previous equation

r/ro=(1+γΔT)^(1/3)

r/r0=(1-3.2x10-5(3000-300))^(1/3)=

r/ro=0.9703=97.03%

8 0
3 years ago
Radio waves are the only type of electromagnetic wave that is a transverse wave.
Pavlova-9 [17]

Answer:

false

Explanation:

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Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is abo
hoa [83]

To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

f = 4.11 *10^{12} Hz

A = 1.23 * 10^{-11}m

The angular velocity of a body can be described as a function of frequency as

\omega = 2\pi f

\omega = 2\pi 4.11 *10^{12}

\omega=2.582*10^{13} rad/s

PART A) The expression for the maximum angular velocity is given by the amplitude so that

V = A\omega

V =( 1.23 * 10^{-11})(2.582*10^{13})

V =  = 317.586m/s

PART B) The maximum acceleration on your part would be given by the expression

a = A \omega^2

a =( 1.23 * 10^{-11})(2.582*10^{13})^2

a= 8.2*10^{15}m/s^2

4 0
3 years ago
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
2 years ago
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