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Ierofanga [76]
1 year ago
10

a projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. what

is the projectile's horizontal speed after it has fallen 25 meters? 22 m/s 35 m/s
Physics
2 answers:
STatiana [176]1 year ago
5 0

Answer: 35m/s

Explanation: answer from castle learning

QveST [7]1 year ago
3 0

A projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. projectile's horizontal speed after it has fallen 25 meters is 35 m/s.

<h3>what is speed ?</h3>

Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

The formula of speed is represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

There are four types of speed such as Uniform speed, Variable speed, Average speed, Instantaneous speed

Uniform speed is an uniform speed when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the uniform speed the ratio of total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as an object is moving with variable speed, then the speed at any instant of time is known as instantaneous speed.

For more details regarding speed, visit

brainly.com/question/13263542

#SPJ2

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A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
Which of the following statements would NOT result in an eclipse? a. The Earth is positioned directly between the Sun and the Mo
vovangra [49]

c. The Moon is positioned directly between the earth and the sun is the statement that does not result in an eclipse.

Explanation:

  • The Sun is completely blocked in a solar eclipse because the Moon passes between Earth and the Sun.
  • it is just the right distance away from Earth, the Moon can fully blocks the Sun's light from Earth's perspective
3 0
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A hammer strikes one end of a thick iron rail of length 5.60 m. A microphone located at the opposite end of the rail detects two
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Answer:

A) The wave that travels through the rail reaches the microphone first.

B) separation in time between the arrivals of the two pulses is 0.01539 seconds.

Explanation:

Detailed explanation and calculation is shown in the image below

4 0
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What are the differences and the similarities between psychology and the social science
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Answer:

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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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