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Eddi Din [679]
3 years ago
14

A car moves along a straight path with a speed of 30kmh. How much distance will the car travel in 20 minutes​

Physics
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

speed \:  =  \frac{distance}{time}  \\ speed \times time = distance \\ 8.3ms^{ - 1} \times 1200s \\ 9960m

30km/h = 8.3ms^-1

20min = 1200s

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Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s
umka2103 [35]
  • initial velocity=u=0m/s
  • Final velocity=v=4m/s
  • Time=t=2.5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}

\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}

\\ \sf\longmapsto Acceleration=1.6m/s^2

8 0
3 years ago
What is the speed of an object at rest
sammy [17]
The speed of an object is measured by calculating the distance it travels per unit time.
The formula in Physics for calculating speed is = distance/ time.

Since an object at rest is not covering any distance, the speed is always assumed to be 0 m/s.
4 0
3 years ago
If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
When are magnets attracted and when are they repelled.
vova2212 [387]
Magnets are attracted when each of the different sides, most commonly known as "North" and "South", are facing each other. They repel when North and North, or South and South are facing each other.
3 0
3 years ago
Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a
Novosadov [1.4K]

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

        1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

        1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

6 0
3 years ago
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