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3 years ago

The last multiple choice answer is

Physics

1 answer:

N76 [4]3 years ago

5 0

Answer:

Answer: C

Explanation:

The closer the city is by the ocean, the cooler it will get because of the winds it is created from the ocean. Meanwhile, Atlanta is slightly away from the ocean but wont be as cool as the Tybee Island. So the heat will hit more to the city away from the ocean.

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What happens when light passes through a prism or diffraction grating?

Dimas [21]

Answer:

The light scatters in all directions.

6 0

3 years ago

While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential

BabaBlast [244]

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

8 0

3 years ago

What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical posi

maksim [4K]

Answer:

ΔU = 2 mg h

Explanation:

In a spring mass system the potential energy is U = m g h

where h is measured from the equilibrium point of the spring

the potential energy at the highest point is

U₁ = m g h

the potential energy at the lowest point is

U₂ = m g (-h)

instead in this energy it is

ΔU = 2 mg h

In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero

4 0

3 years ago

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the

Novay_Z [31]

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

sin θ = y / L

we will then substitute

y’ / L = y/L 1/2

y' = y / 2

this means that when the slit width is doubled the pattern width will then be halved

4 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago