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3 years ago

12

A baseball player throws a baseball. The ball traveled 35 yards before it landed on the ground. The ball stayed in the air for 3

.8 seconds. What is the average speed of the ball?

2 answers:

Ann [662]3 years ago

6 0

(35 yards) / (3.8 seconds) = 9.21 yards per second

Free_Kalibri [48]3 years ago

4 0

Average speed = (distance traveled) / (time to cover the distance)

= (35 yards) / (3.8 seconds)

= 9.21 yards per second

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

3 years ago

Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-

n200080 [17]

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$

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4 years ago

The force measuring instrument is called

Phoenix [80]

This instrument is called a spring scale.

5 0

3 years ago

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What planet has a visible mark referred to as the the Great Red Spot?

vivado [14]

Jupiter that is the answer

good luck

6 0

3 years ago