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3 years ago

Which is more reliable—using a manual stop watch or using light gates? Explain.

1 answer:

7 0

Light gates are more reliable. When using a manual stop watch, it is difficult to stop it at an exact time. A light gate is able to detect when an object passes through a 'gate' with the infrared transmitter and receiver.

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HELP These images represent waves. Which wave has the highest frequency?

Ierofanga [76]

Wave C

because it is the closest or each wave is closer than the others

7 0

3 years ago

Read 2 more answers

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

3 years ago

Someone please help.

Anettt [7]

Answer:

B is the answer

Explanation:

7 0

2 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of 6.45 m left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago

In deep space, sphere A of mass MA is located at the origin of an x axis and sphere B of mass MB is located on the axis at a poi

Nookie1986 [14]

Answer:

The gravitational potential energy of the two-sphere system just as B is released is

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Explanation:

The gravitational potential energy of two masses (m and M), separated by a distance, d, is given as

U = -(GMm/d)

For our question,

Mass of object 1 = MA

Mass of object 2 = MB

Distance between them = x₁

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Hope this Helps!!!

6 0

3 years ago