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A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Answers

Answer 1

Answer:

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^(\circ)$

Let u be the initial speed of ball

Range $=(u^2\sin 2\theta )/(g)$

$167=(u^2\sin (66.2))/(9.8)$

$u^2=1788.71$

$u=√(1788.71)$

$u=42.29 m/s$

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Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

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Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)

Answers

Answer:

Acceleration = 8.27 cm/s²

Explanation:

We are given;

initial velocity; v_i = 10.5 cm/s

Initial position; x_i = 2.72 cm

Time; t = 2.30 s

final position; x_f = 5.00 cm

To find the acceleration, we will make use of the formula;

x_f - x_i = (v_i * t) - (½at²)

Plugging in the relevant values, we have;

5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)

2.28 = 24.15 - 2.645a

24.15 - 2.28 = 2.645a

2.645a = 21.87

a = 21.87/2.645

a = 8.27 cm/s²

Using the kinematic equation, the acceleration of the object was calculated to be approximately8.27 cm/s² given its initial velocity, position, time, and final position.

We are given:

Initial velocity (vᵢ) = 10.5 cm/s

Initial position (xᵢ) = 2.72 cm

Time (t) = 2.30 seconds

Final position ( $x_f$ ) = 5.00 cm

We want to find the acceleration (a) of the object using the kinematic equation:

x₋ᵢ - xᵢ = (vᵢ * t) - (1/2) * a * t²

Now, let's substitute the given values:

5.00 cm - 2.72 cm = (10.5 cm/s * 2.30 s) - (1/2) * a * (2.30 s)²

Simplify the equation:

2.28 cm = 24.15 cm - (1/2) * a * 5.29 s²

Now, isolate 'a' by rearranging the equation:

-1.09 cm = (-1/2) * a * 5.29 s²

To remove the negative sign, multiply both sides by -1:

1.09 cm = (1/2) * a * 5.29 s²

Next, solve for 'a' by multiplying both sides by (2 / 5.29):

a ≈ (1.09 cm) / (2 / 5.29) s²

a ≈ 8.27 cm/s²

So, the acceleration of the object is approximately 8.27 cm/s².

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A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Much of our knowledge of the interior of the Earth comes from the study of planetary vibrations, which is the science of

Answers

Answer:

Seismology.

Explanation:

Seismology is the beach of physical science that deals with the study of vibrations that comes out from the interior of the earth onto the surface and these vibrations are in the form of seismic waves that are primary, secondary, and surface waves.

The science of seismology tells about the magnitude and intensity of these waves that lead to planetary vibrations. These waves trigger earthquakes, floods, and even landslides.

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m

Answers

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u = 3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

v² = u² + 2as

0² = 2.98² + 2 x -9.81 x s

s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

You received a shipment 20 days ago of 13 I for treatment of hyperthyroidism. What fraction of the original shipment would you still have with a half-life of 8.040 days for 31?