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OverLord2011 [107]
3 years ago
14

To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

ossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 31◦
from the horizontal.
The acceleration of gravity is 9.8 m/s^2.
By how much vertical distance does the ball clear the crossbar?
Answer in units of m.
Physics
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

u_{x}=u\cos\theta

Put the value into the formula

u_{x}=24\cos(31)

u_{x}=20.5\ m/s

We need to calculate the vertical velocity

Using formula of vertical velocity

u_{y}=u\sin\theta

Put the value into the formula

u_{y}=24\sin(31)

u_{y}=12.3\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{u_{x}}

Put the value into the formula

t=\dfrac{44}{20.5}

t=2.1\ sec

We need to calculate the vertical height

Using equation of motion

h=u_{y}t+\dfrac{1}{2}at^2

Put the value into the formula

h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2

h=4.2\ m

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

d=h-3.05

Put the value of h

d=4.2-3.05

d=1.15\ m

Hence, The distance by the ball clear the crossbar is 1.15 m

You might be interested in
How much work is done on a satellite in a circular orbit about earth?
alexgriva [62]

Answer:

0 J

Explanation:

W = Work done on the satellite in circular orbit about earth by earth

F = Force on gravity on satellite by earth

d = displacement of the satellite

\theta = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J

3 0
2 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

Q_H

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

Q_C>0

That is why the answer will be

Q_H ,Q_C>0

8 0
3 years ago
car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw
sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = m \times  v_f - m \times  v_0

Where;

v_f = The final velocity

v_0 = The initial velocity

For the the 265 kg mass, we have;

v_f = 0.1627 m/s

v_0 = 1.5 m/s

Which gives the impulse a s F×Δt =  265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0
3 years ago
A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of  output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( \frac{1}{2fRC})V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute  for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 /  ( (120 * 1 * 62.2) )15V

= 2V

The peak-to-peak ripple voltage = 2V

3 0
3 years ago
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