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Nataly_w [17]
3 years ago
7

A projectile is launched vertically at 100 m/s. If air resistance can be ignored, at what speed will it return to its initial le

vel?
Physics
2 answers:
eduard3 years ago
7 0
<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

madreJ [45]3 years ago
6 0

Answer:

100 m/s

Explanation:

given,

vertical velocity of the projectile = 100 m/s

When the Projectile is launched vertically it will have two-component one is the vertical component and another is a horizontal component.

The horizontal component of velocity will remain constant.

The vertical component of the velocity will be zero at the highest point is zero and the second half motion of the projectile is the mirror of the first half movement of the projectile.

When the object reaches the initial level the velocity will be the same i.e. 100 m/s.

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A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec
Degger [83]

Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.

H.The velocity vector of the ball changes as it travels through the air.

Explanation:

As the ball is projected horizontally so here the vertical component of the velocity is zero

So the time to reach the ground is given as

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

so we will have

t = \sqrt{\frac{2H}{g}}

so this is the same time as the ball is dropped from H height

Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.

So overall the velocity vector will change due to net acceleration g

4 0
3 years ago
A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.
harina [27]

I can not solve the problem if I do not have the mass.

3 0
3 years ago
It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0
VikaD [51]
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have 

rate = \frac{-(C_{new} - C_{old})}{t}

where C are the concentration values and t is the time taken for it to decompose.

14 = \frac{-(0.02 - 0.06)}{t}
t = 0.003 s

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0
3 years ago
The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,
ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy =  -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

dt/dy = -54× 2y / (7 + x² + y²)²      .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

7 0
3 years ago
Can someone please help me with this? <br>​
Sauron [17]

Divide 56 by 60 then x the answer by 6 then you get 5.6km which is your answer.

5 0
3 years ago
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