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3 years ago

A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

igh (in m) was the building?

Physics

1 answer:

Aloiza [94]3 years ago

6 0

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$

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A bullet of mass 11.1 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.01 kg,

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Answer:

a) The initial speed of the bullet is 488 m/s

b) The loss of kinetic energy is 1.3 × 10³ J.

Explanation:

Hi there!

To solve this problem we have to use the conservation of momentum:

initial momentum of the bullet + initial momentum of the block =

final momentum of the block-bullet system

The momentum of an object is calculated as follows:

p = m · v

Where:

p = momentum

m = mass of the object.

v = velocity.

Then, in our system:

p₁₁ = initial momentum of the bullet.

p₂₁ = initial momentum of the block.

p₃₂ = final momentum of the block-bullet system.

p₁₁ + p₂₁ = p₃₂

The initial momentum of the bullet will be:

p₁₁ = m · v

p₁₁ = 0.0111 kg · v

The initial momentum of the block will be:

p₂₁ = 1.01 kg · 0 m/s = 0 kg · m/s

The final momentum of the block-bullet system will be:

p₃₂ = (1.01 kg + 0.0111 kg) · 5.30 m/s

Then, by conservation of the momentum:

initial momentum of the bullet = momentum of the block-bullet system

0.0111 kg · v = (1.01 kg + 0.0111 kg) · 5.30 m/s

v = ((1.01 kg + 0.0111 kg) · 5.30 m/s)/ 0.0111 kg

v = 488 m/s

The initial speed of the bullet is 488 m/s

b) The initial kinetic energy (KE) of the system is the kinetic energy of the bullet because the block is at rest:

KE = 1/2 · m · v²

KE = 1/2 · 0.0111 kg · (488 m/s)²

KE = 1.32 × 10³ J

The final kinetic energy of the system will be the kinetic energy of the block-bullet system:

KE = 1/2 · (1.01 kg + 0.0111 kg) · (5.30 m/s)²

KE = 14.3 J

The loss of kinetic energy will be:

initial kinetic energy - final kinetic energy

1.32 × 10³ J - 14.3 J = 1.3 × 10³ J

The loss of kinetic energy is 1.3 × 10³ J.

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3 years ago

El. A horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. How much work is done by

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The same formula of work can be applied to all the questions. The answers are:

E1. 100J

E2. 40N

E3. 5m

E4. a.) 360 J b.) 240 J c.) 120 J

El. If a horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. The formula to get the much work that will be done by the 40 - N force will be

Work done = force x distance

Work done = 40 x 2.5

Work done = 100 J

E2. If a woman does 160 J of work to move a table 4 m across the floor. We will use the same formula to calculate the magnitude of the force that the woman applied to the table assuming the force is applied in the horizontal direction.

Work done = force x distance

160 = 4F

F = 160/4

Force F = 40 N

E3. Given that a force of 60 N used to push a chair across a room does 300 J of work. Same formula to get how far the chair move in this process.

Work done = force x distance

300 = 60 x distance

distance = 300/60

Distance = 5 m

Therefore, the chair moved 5m away.

E4. Given that a rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. And a frictional force of 120 N opposes this motion.

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W = F x S

W = 180 x 2

W = 360 J

b. What is the work done by the frictional force?

W = $F_{r}$ x s

W = 120 x 2

W = 240 J

c. What is the total work done on the crate?

W = (F - $F_{r}$ ) x distance

Where $F_{r}$ = frictional force

Substitute all the parameters

W = (180 - 120) x 2

W = 60 x 2

W = 120 J

Learn more about work here: brainly.com/question/8119756

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2 years ago

Astronomers use the _______ of stars to determine how hot they are and their _______ to estimate how far away they are. A. brigh

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Astronomers use the color of stars to determine how hot they are and their brightness to estimate how far away they are. <em>(D)</em>

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A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

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To solve this exercise it is necessary to use the concepts related to Difference in Phase.

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From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

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3 years ago

A 5. 3 ft -ft-tall girl stands on level ground. The sun is 30 ∘ above the horizon. How long is her shadow?

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The trigonometric ratios are sine, cosine and tangent. We can use cosine to find the length of the shadow from the calculation as 1.7 ft.

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The term trigonometric ratio has to do with sine, cosine and tangent which are used to solve problems that involve the right angled triangle.

Using the geometry of the right angle triangle, we can find the tangent of the angle 30 in order to obtain the length of the shadow.

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2 years ago