PHYSICS COLLEGE

18. Wind speed on Earth is reduced by which?A. temperature
B. friction
C. weather
D. convergence

Answers

Answer 1

Answer:

Answer:

Temperaturereduces the wind speed on earth.

Explanation:

Wind speed is an atmospheric quantity that causes the wind to flow from high to low level due to a change in temperature.
Wind speed is controlled by the pressure gradient factor and it increases the wind speed with increase in pressure gradient.
Wind speed decreases with an increase in temperature value as it absorbs the heat developed in the system and wind gets slowed down.
Wind speed is measured in velocity and anemometer is used to measure the rate of wind speed.

Answer 2

Answer:

Answer:

your answer would be A. temperature

Explanation:

Wind speed on Earth is reduced by temperature, it depends on the temperature because let's say there is a nice sunny day, then obviously it will reduce the wind speed

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An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

A baseball leaves the bat with a speed of 40 m/s at an angle of 35 degrees. A 12m tall fence is placed 130 m from the point the ball was struck. Assuming the batter hit the ball 1m above ground level, does the ball go over the fence? If not, how does the ball hit the fence? If yes, how far beyond the fence does the ball land?

Answers

Answer:

The ball land at 3.00 m.

Explanation:

Given that,

Speed = 40 m/s

Angle = 35°

Height h = 1 m

Height of fence h'= 12 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$V_(x)=V_(i)\cos\theta$

$V_(x)=40*\cos35$

$V_(x)=32.76\ m/s$

We need to calculate the time

Using formula of time

$t = (d)/(v)$

$t=(130)/(32.76)$

$t=3.96\ sec$

We need to calculate the vertical velocity

$v_(y)=v_(y)\sin\theta$

$v_(y)=40*\sin35$

$v_(y)=22.94\ m/s$

We need to calculate the vertical position

Using formula of distance

$y(t)=y_(0)+V_(i)t+(1)/(2)gt^2$

Put the value into the formula

$y(3.96)=1+22.94*3.96+(1)/(2)*(-9.8)*(3.96)^2$

$y(3.96)=15.00\ m$

We need to calculate the distance

$s = y-h'$

$s=15.00-12$

$s=3.00\ m$

Hence, The ball land at 3.00 m.

A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

Answers

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

Final answer:

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.

Explanation:

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.

After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

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Find the force on a 5 pС charge in a place where the electric field is 400 N/C.

Answers

Answer:

Electric force, $F=2* 10^(-9)\ N$

Explanation:

It is given that,

Charge on the particle, $q=5\ pC=5* 10^(-12)\ C$

Electric field, $E=400\ N/C$

Let F is the electric force acting on the charged particle. The electric force per unit electric charge is called electric field. Mathematically, it is given by :

$F=qE$

$F=5* 10^(-12)\ C* 400\ N/C$

$F=2* 10^(-9)\ N$

So, the force acting on the charged particle is $2* 10^(-9)\ N$ . Hence, this is the required solution.

In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)

Answers

Answer:

a. $12.3m/s^(2)$

b. $21.93m/s^(2)$

Explanation:

From the data given, the radius is 5.0m, and the time taken to complete one circle is 4.0secs

Since the motion is in a circular part, we can conclude that the total distance covered in this time is given as circumference of the circle.

which is expressed as

$Distance=2\pi R$

To determine the speed, we use the equation

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(4)\n Speed=7.85m/s$

The acceleration as required is expressed as

$a=(v^(2))/(r)\n a=(7.85^(2))/(5)\n a=12.3m/s^(2)$

if the speed increase and it takes 3secs to complete one circle, the speed is

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(3)\n Speed=10.47m/s$

and the acceleration becomes

$a=(v^(2))/(r)\n a=(10.47^(2))/(5)\n a=21.93m/s^(2)$

The acceleration of the passengers in the vertical circle carnival ride is 19.6 m/s^2. When the time taken to complete one circle is 3.0 s, the new acceleration is 26.13 m/s^2.

The acceleration of the passengers can be determined using the centripetal acceleration formula, which is given by a = v^2 / r.

In this case, the velocity v can be found by dividing the circumference of the circle (2πr) by the time taken to complete one circle (T). The radius r is given as 5.0 m. Plugging in the values, we have:

a = (v^2) / r = ((2πr / T)^2) / r = (4π^2r) / T^2 = (4π^2 * 5.0) / 16.0 = 19.6 m/s^2

To find the new acceleration when the time taken to complete one circle is 3.0 s, we can use the proportional reasoning to determine the relationship between the two accelerations. Since the time is inversely proportional to the acceleration, when T is 3.0 s, the new acceleration arad can be found using the equation:

arad / 19.6 = 4.0 / 3.0

Simplifying the equation, arad = (19.6 * 4.0) / 3.0 = 26.13 m/s^2

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PLEASE HELP QUICKLY!A volleyball player jumps to hit a ball horizontally at 7.0 m/s straight on. If the
height at which the ball was hit is 3.0 m tall, how far did the ball go horizontally
before it hit the ground?
5.5 m
3.6 m
O 4.3 m
4.2 m