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2 years ago

What property of matter changes depending on location?

Physics

1 answer:

Angelina_Jolie [31]2 years ago

8 0

Answer:

weight. The force with which an object near the Earth or another celestial body is attracted toward the center of the body by gravity. An object's weight depends on its mass and the strength of the gravitational pull.

Explanation:

This may be right i'm not 100%

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A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the

wel

Answers:

B.) $F cos\theta=F_{n}$

C.) $F sin\theta=F_{g} \pm F_{f}$

Explanation:

The image attached shows the way the force $F$ is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

$-F_{n}+F cos\theta=0$ (1)

Where:

$F_{n}$ is the Normal force

$F$ is the magnitude of the force exerted on the block

$\theta$ is the angle

Net Force in Y:

$F sin\theta \pm F_{f}-F_{g}=0$ (2)

Where:

$F_{f}$ is the Friction force (it is expresed with the $\pm$ sign because this force may be up or down, we cannot know because the block is at rest)

$F_{g}$ is the gravity force

Rewrittin (1):

$F cos\theta=F_{n}$ (3) This is according to option B

Rewritting (2):

$F sin\theta=F_{g}\pm F_{f}$ (3) This is according to option C

3 0

2 years ago

A 30.0-Î¼F capacitor is connected to a 49.0-Î© resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms

Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

$Z=\sqrt{R^2+X_{c}^2}$ .....(I)

We need to calculate the value of $X_{c}$

Using formula of $X_{c}$

$X_{c}=\dfrac{1}{2\pi f c}$

$X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}$

$X_{c}=88.42\ \Omega$

Put the value of $X_{c}$ into the formula of impedance

$Z=\sqrt{(49.0)^2+(88.42)^2}$

$Z=101.08\ \Omega$

(a). We need to calculate the rms current in the circuit

Using formula of rms current

$I_{rms}=\dfrac{V}{Z}$

$I_{rms}=\dfrac{30.0}{101.08}$

$I_{rms}=0.30\ A$

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

Put the value into the formula

$V_{rms}=0.30\times49.0$

$V_{rms}=14.7\ V$

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.30\times88.42$

$V_{rms}=26.53\ V$

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0

2 years ago

Which statement is true according to Newton's first law of motion?

katrin2010 [14]

C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

hope this helps and have a great day :)

8 0

3 years ago

Read 2 more answers

At what point in the swing of the pendulum is the potential energy completely converted into Kinetic energy

artcher [175]

Point a because point a is the highest at potential energy converting into the highest kinetic energy.

4 0

2 years ago

Read 2 more answers

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

$d=vt=(16 m/s)(2 s)=32 m$

So, the closest answer is B.

5 0

3 years ago