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gayaneshka [121]

3 years ago

5

Which image depicts projectile motion?

2 answers:

Elis [28]3 years ago

6 0

Answer:

the answer is the second one

Explanation:

the person is moving through the air while the other people are either not moving forward or backwards or on the ground with full controll over where they go

tia_tia [17]3 years ago

4 0

Answer:

Second picture is representing projectile motion

Explanation:

As we know that in projectile motion one of the component of the motion of object will remains constant while other component which is vertical in direction will change due to gravity.

So here in this type of motion the path of the object is same as the parabolic path.

Here we know that vertical component of the motion will change due to gravity while horizontal component will remain constant always

So here we can say that

correct image is second image

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dezoksy [38]

The correct answer would be 40% plz mark brainliest!

8 0

3 years ago

This graph shows how an object’s speed changes over time.

Digiron [165]

Answer:

c. The forces acting on the box are balanced.

3 0

2 years ago

1. °C= (°F-32) X 5/9. what will be the answer?

aleksley [76]

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

Take 25°C

$\\ \sf\longmapsto \dfrac{9}{5}(25)+32$

$\\ \sf\longmapsto 9(5)+32$

$\\ \sf\longmapsto 45+32$

$\\ \sf\longmapsto 77°F$

3 0

3 years ago

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

where

F is the magnitude of the force

d is the distance covered by the object

$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

$\theta=0$

And the distance covered is equal to the circumference of the circle, which is:

$d=2\pi r=2\pi (2.5 m)=15.7 m$

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago

A cabbie is trying to stop when he notices a fare is whistling them over. The

liberstina [14]

K.E=18750J
Mass=m=2100kg
Velocity=v

$\boxed{\sf K.E=\dfrac{1}{2}mv^2}$

$\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2$

$\\ \sf\longmapsto 18750=1050v^2$

$\\ \sf\longmapsto v^2=\dfrac{18750}{1050}$

$\\ \sf\longmapsto v^2=17.85m^2$

$\\ \sf\longmapsto v=\sqrt{17.85}$

$\\ \sf\longmapsto v=4.1m/s$

7 0

3 years ago