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3 years ago

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A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (A)

Its velocity is not zero, but its acceleration is zero. (B) Its velocity and its acceleration are both zero. (C) Its velocity is perpendicular to its acceleration. (D) Its acceleration depends on the angle at which the ball was thrown. (E) None of the above statements are true.

1 answer:

11111nata11111 [884]3 years ago

5 0

Answer:

C. Its velocity is perpendicular to its acceleration

Explanation:

Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude

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I dont get why its c i dont under the graph for the vertical component in option c

jek_recluse [69]

The vertical component grows at an increasing rate because
gravity is ACCELERATING the object vertically. So its speed
keeps growing. Speed is the slope of the displacement graph.

7 0

3 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

So

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

So

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

3 years ago

At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at

Agata [3.3K]

We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.

8 0

3 years ago

Read 2 more answers

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

Agata [3.3K]

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

Explanation:

For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

$\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m$

Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

$\boxed{T = 1.5*10^{-6}N}$

or in micro newtons

$T =1.5\mu N$

4 0

3 years ago

Energy Transformation

Serjik [45]

Answer:

<em>K =400000 J</em>

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The car has a mass of m=2000 Kg and travels at v=20 m/s. Calculating the kinetic energy:

$\displaystyle K=\frac{1}{2}2000*20^2$

Calculating:

K =400000 J

3 0

3 years ago