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Arada [10]
2 years ago
14

A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sle

d attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Physics
1 answer:
ycow [4]2 years ago
4 0

Answer:

\mu=0.185

Explanation:

From the question we are told that:

Mass m=17kg

Force F=33N

Velocity v=1.6m/s

Distance d= 9.8m

Generally the equation for Work done is mathematically given by

 W=\triangle K.E+\triangle P.E

Where

 \triangle K.E=(F-F_f)*2

 F_f=F+\frac{\triangle K.E}{d}

 F_f=33+\frac{0.5*17*1.6^2}{9.8}

 F_f=30.8N

Since

 f = \mu*m*g

 \mu= 30.8/(m*g)

 \mu= 30.8/(17*9.81)

 \mu=0.185

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A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0
3 years ago
Read 2 more answers
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr
fredd [130]

Answer:

Frequency, f=2.18\times 10^9\ Hz

Explanation:

We have,

Speed of radio waves is 3\times 10^8\ m/s

Wavelength of radio waves is \lambda=0.137\ m

It is required to find the frequency of the radio waves. The speed of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz

So, the frequency of the radio wave is 2.18\times 10^9\ Hz.

8 0
2 years ago
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0
1 year ago
We would be more likely to get an accurate measurement of the distance from Earth to a nearby star if we took the angular measur
NeTakaya

Answer:

6 month interval

Explanation:

The distance to a nearby star in theory is more simple than

one might think! First we must learn about the parallax effect. This is the mechanism our eyes use to perceive things at a distance! When we look at the star from the earth we see it at different angles throughout the earth's movement around the sun similar to how we see when we cover on eye at a time. Modern telescopes and technology can help calculate the angle of the star to the earth with just two measurements (attached photo!) Since we know the distance of the earth from the sun we can use a simple trigonometric function to calculate the distance to the star. The two measurements needed to calculate the angle of the star to the earth caused by parallax (in short angle θ) are shown in the second attached photo.

So using a simple trigonometric function Sin\theta=\frac{r}{d} we can solve for d which is the distance of the earth to the star:

d=\frac{r}{Sin\theta}

In the first attached photo a picture where r is the distance to the star and the base of the triangle is the diameter of the earth.

8 0
3 years ago
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