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3 years ago

⚠️I need help with the last question!⚠️

Physics

1 answer:

Ket [755]3 years ago

5 0

Answer:

I can't do your work for you but I can explain the last question;

The want you to tell them (In at least 3 sentences) Why you think your answers are correct or how your answer's match your hypothesis.

A hypothesis (for a little more help) is an idea or explanation that you then test through study and experimentation.

Outside science, a theory or guess can also be called a hypothesis. A hypothesis is something more than a wild guess but less than a well-established theory. Anyone who uses the word hypothesis is making a guess.

Explanation:

Sorry I didnt give you the exact answer but I hope this helps :)

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What do we call the movement of electric charge?<br> A. Power B. Voltage C. Resistance D. Current

Alex Ar [27]

D. Current

Hope I helped!

4 0

3 years ago

Read 2 more answers

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

$\lambda = \frac{v}{f}$

Where,

$\lambda =$ Wavelength

f = Frequency

v = Velocity

Our values are given as,

$f = 2.8*10^3Hz$

$v = 340m/s \rightarrow$ Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

$\lambda = \frac{340}{2.8*10^3}$

$\lambda = 0.1214m$

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

A car is traveling in uniform circular motion on a section of flat roadway whose radius is . The road is slippery, and the car i

Alona [7]

Answer:

The smallest radius will be four (4) times the initial radius

Explanation:

The car maintains a constant angular speed. According to Newton's Second Law F = m a

1. $F_{r}=m*A_{n}$

2. $A_{n}=\frac{v^2}{R_{p}}$

Replacing 2 in 1

3. $F_{r}=m*\frac{v^2}{R_{p}}$

Where:

Fr= Frictional force

Rp= Initial Radius

An= Centripetal Acceleration

M= Mass

V= Velocity

Also we have that:

4. $F_{r}=\mu *W=\mu*m*g$

μ= Coefficient of friction between the car and the surface

M= Mass

W= Weight

G= Gravity

r is cleared from equation 3

5. $R_{p}=m*\frac{v^2}{F_{r}}$

Replacing 4 in 5

6. $R_{p}=m*\frac{v^2}{\mu*m*g}$

Simplifying

7. $R_{p}=\frac{v^2}{\mu*g}$

Now we have a new velocity equal to twice the initial velocity, We replace it by 2v in equation 7

8. $R_{n}=\frac{(2v)^2}{\mu*g}$

Computing

9. $R_{n}=\frac{4v^2}{\mu*g}$

Replacing 5 in 9

$R_{n}=4*R_{p}$

8 0

3 years ago

If you push a crate with a force of 100 N and it slides at a constant speed and in the same direction, how much friction is acti

Nadusha1986 [10]

Answer: 100 N

Explanation: Taking into account the second Newton Law the total force applied to any system is equal to the mass *acceleration.In this case the crate moves at constant speed so the accelaration is zero. In order to satisfy this fact, the friction force must be equal the applied force of 100 N .

8 0

3 years ago