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Allisa [31]
2 years ago
6

A freight train car is moving at a constant speed with respect to the ground of Vcg. A man standing on a flatbed car throws a ba

ll into the air and catches it as it falls. Relative to the flatbed car, the initial velocity of the ball is Vibc straight up. a. What are the magnitude and direction of the initial velocity of the ball as seen by a second man standing on the ground next to the track
Physics
1 answer:
11111nata11111 [884]2 years ago
5 0

Answer:

chiicken

Explanation:

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QUICK HELP PLEASE!
IgorLugansk [536]

Answer:

60/90

Explanation:

I think because the train's highest velocity is 60 n the time is 90

5 0
3 years ago
A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the
xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2

\hat{ECD} = tan^{-1} 2 = 63.43^o

This is also the angle ACB, so we can find the length AB:

tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}

2 = \frac{AB}{234}

AB = 2*234 = 468 m

So the height of the building is 468m

5 0
3 years ago
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

5 0
2 years ago
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     v=\sqrt{\frac{T}{(m/l)}}

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

(m/l)=\frac{10}{V^2}

Explanation:

From the question we are told that

Speed of a transverse wave given by

v=\sqrt{\frac{T}{(m/l)}}

Maximum Tension is T=10.0N

Generally making (m/l) subject from the equation mathematically we have

v=\sqrt{\frac{T}{(m/l)}}

v^2=\frac{T}{(m/l)}

(m/l)=\frac{T}{V^2}

(m/l)=\frac{10}{V^2}

Therefore the Linear mass in terms of Velocity is given by

(m/l)=\frac{10}{V^2}

8 0
3 years ago
A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

4 0
3 years ago
Read 2 more answers
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