What variable is represented on the y-axis?

Study the velocity vs. time graph shown.

Sindrei [870]

Answer:24m

Explanation:

5 0

3 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

3 years ago

The skateboarder starts at the top of the ramps, and rolls down and back up the other side. Which graph shows the most Kinetic E

DaniilM [7]

Answer:

C

Explanation:

Usually when you are at the bottom you are at peak speed. It also shows that Kinetic Energy is the green bar and in picture C the green bar is highest.

6 0

2 years ago

Gravitational forces is ------ force

mel-nik [20]

Answer:

Gravitational force is <u>noncontact</u> force

Explanation:

Contact force occurs due to the contact between two different objects. Non-contact force occurs due to either attraction or repulsion between two objects such that there is no contact between these objects. There is no field linked with the contact force. ... Gravitational force is an example of a non-contact force.

7 0

3 years ago

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

3 years ago