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3 years ago

What is the net force on this object: F norm = 20N F frict = 14N F app = 22N F grav = 20N?

Physics

2 answers:

sergey [27]3 years ago

8 0

Answer:

8N is the net force on the object.

Explanation:

Given that,

$\mathrm{F}_{\text {norm }}$ =20N is normal force acting upward (+20N)

$F_{\text {frict }}$ =14N is frictional force which is opposite to the applied force (-14N)

$\mathrm{F}_{\mathrm{app}}=$ 22N is applied force which is applied on right side (+22N)

$\mathrm{F}_{\mathrm{grav}}$ =20N is gravitational force acting downward (-20N)

We know that ,

$\text { Fnet }$ =applied force $\left(\mathrm{F}_{\mathrm{app}}\right)$ +frictional force $\left(F_{\text {frict }}\right)$ +normal force $\left(\mathrm{F}_{\mathrm{norm}}\right)$ +gravitational force $\left(F_{\text {grav}}\right)$

$\mathrm{F}_{\mathrm{net}}=$ 22+(-14)+ 20+(-20)

$\mathrm{F}_{\mathrm{net}}=$ 22-14+20-20

$\mathrm{F}_{\mathrm{net}}=8 \mathrm{N}$

Therefore the net force of the object be 8N.

viva [34]3 years ago

4 0

Answer:

8N i believe

Explanation:

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3 years ago

A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

harina [27]

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

7 0

2 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

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2 years ago

A bus driver drove her bus for 5 miles. She drove that distance in a total of 30 minutes. What was the truck driver's average sp

nexus9112 [7]

Answer:

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