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Natalka [10]
3 years ago
13

g John is walking along a trail when he comes to the bottom of a steep cliff. Before trying to climb up it, he wonders how high

it is. Jake happens to have a bow and arrow with him, and he knows it is able to launch an arrow with a maximum speed of 60.0 m/s. After shooting the arrow straight up with the bow fully extended, he hears it hit the ground at the top of the cliff 7.00 s after he launches it. How high is the cliff
Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

179.655m

Explanation:

Given

Maximum speed of the arrow v = 60m/s

Time taken to hit the top of the cliff t = 7.0s

Required

Height of the cliff H

Using the equation of motion

H = vt + 1/2gt²

Substitute into the formula:

H = 60(7) + 1/2 (-9.81)(7²) (g is negative due to upward motion of the arrow)

H = 420-4.905(49)

H = 420-240.345

H = 179.655m

Hence the cliff is 179.655m high

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zhuklara [117]

what do you need help with can i help with something or do you what me to give you the answer ask me anething and i got youExplanation:

8 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
3 years ago
A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?
lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos\theta, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, \theta = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0
2 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
3 years ago
An organized arrangement of information in labeled rows and columns is called a _____?
Sav [38]
It is called a table
7 0
3 years ago
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