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Degger [83]
4 years ago
5

Which of these is the largest?

Physics
1 answer:
seropon [69]4 years ago
7 0
The answer is b meteoroid
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The magnetic field is decreasing. What is the induced electric field?
melisa1 [442]

Answer: The induced electric field will be constant i.e zero

Explanation:

Magnetic field is a region of space where magnetic force can be felt while electric field is a region of space where electric force can be felt. This two field are related because they tend to talk about force of attraction and repulsion. Example is running a current through a thin wire near a magnet, the direction of one of the field will determine the direction of the other.

According to the question, if the magnetic field is decreasing, the electric field will adjust by traveling in opposite direction to the magnetic field just to make the field constant.

5 0
3 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
A Chinook salmon can jump out of water with a speed of 6.70 m/s6.70 m/s . How far horizontally dd can a Chinook salmon travel th
aksik [14]

Answer:

R = 3.88 m

Explanation:

As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:

V₀ = Launch Speed = 6.7 m/s

θ = Launch Angle = 29°

R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?

The value of the range of a projectile is given by the following formula:

R = (V₀² Sin 2θ)/g

R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]

R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]

<u>R = 3.88 m</u>

8 0
4 years ago
Help please (about projectile motion at an angle)
PolarNik [594]

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

7 0
3 years ago
A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit
Elis [28]

Answer:

(a) Angular momentum of disk is 1.343\, \frac{kg.m^{2}}{s}

(b) Angular velocity of the disk is 353\frac{rad}{s}

Explanation:

Given

Rotational inertia of the disk , I=3.8\times 10^{-3}kg.m^{2}

Torque delivered by the motor , \tau =17N.m

Torque is applied for duration , \Delta t=79ms=0.079s

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}

=>L=1.343\, \frac{kg.m^{2}}{s}

Thus angular momentum of disk is 1.343\, \frac{kg.m^{2}}{s}

(b)

Since Angular momentum , L=I\omega

where \omega= Angular velocity of the disk

=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}

\therefore \omega =353\frac{rad}{s}

Thus angular velocity of the disk is 353\frac{rad}{s}

5 0
3 years ago
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