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Alinara [238K]
1 year ago
13

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

t what speed does the baseball leave your hand?
Physics
1 answer:
garri49 [273]1 year ago
7 0

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

You might be interested in
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the f
sukhopar [10]

Answer:

c.

Equal to 200 N..........

7 0
1 year ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
2 years ago
A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the
Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

Height, h = 25cm = 0.25 m

Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the  microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure P_r = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

P_r = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

or

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

or

F = 2 × 10⁻³ N

6 0
2 years ago
Read 2 more answers
When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh
Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

5 0
1 year ago
A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses
gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0
1 year ago
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