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Wewaii [24]
3 years ago
10

A bird flew 16 km west in 5 hours , then flew 20 km east in 6 hours . What was the birds velocity?

Physics
1 answer:
LenaWriter [7]3 years ago
5 0
During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)
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A man supports himself and the uniform horizontal beam pulling the rope with a force T.The weights of men and the beam are 883 N
artcher [175]

Answer:

T=502.5N

Ax=171.8N

Explanation:

The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:

vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0

Now  

horizontal forces sum = Ax - Tcos70

Now Moment about B

-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0

Ay=453.6N

Now substitute in sum of vertical forces T=502.5N

Ax=171.8N

3 0
3 years ago
Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?
tensa zangetsu [6.8K]
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3 years ago
Superman is flying 54.5 m/s when he sees
Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations d=v_{0} t+\frac{at^{2} }{2} and v_{f} =v_{0} +at where d is distance, v_{0} is the initial velocity, v_{f} is the final velocity, t is time, and a is aceleration.

Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

Substituting the values:

a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}

To find v_{f} we use the equation v_{f} =v_{0} +at.

Substituting the values:

v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}

5 0
3 years ago
A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

4 0
3 years ago
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
4 years ago
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