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BARSIC [14]
4 years ago
13

A trebuchet launches a pumpkin at an angle of 63 degrees at an initial velocity of 51 m/ s from the ground. What is the range of

the pumpkin?
Physics
1 answer:
Mumz [18]4 years ago
5 0

Answer:

214.72m

Explanation:

The range (R) of a projectile can be calculated using the formula:

R = u²sin2θ/g

Where R = Horizontal range

u = initial velocity

θ = angle of initial velocity

g = acceleration due to gravity (9.8m/s²)

According to the provided information in this question, R= ?, u= 51m/s, θ = 63°, g = 9.8m/s².

Hence, R = u²sin2θ/g

R = 51² × sin 2×63/ 9.8

R = 2601 × sin 126/9.8

R = 2601 × 0.809/9.8

R = 2104.25/9.8

R = 214.72m

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Oduvanchick [21]

Yes, they both cast shadows. A  eclipse occurs when one enters the shadow of the other.

Solar eclipse: Earth in shadow of Moon.

Lunar eclipse: Moon in shadow of Earth.

A solar eclipse is the result of one celestial body being completely or partially obscured by another celestial body along the line of sight of an observer. A solar eclipse occurs when the moon passes between the earth and the sun, obscuring all or part of the sun's disk. A lunar eclipse occurs when the Moon passes through the Earth's shadow, making the Moon's surface completely or partially obscured.

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8 0
2 years ago
An electrician reads a resistance of 65 ohms and when energize the circuit runs 1.77 amps. what voltage to the system run on?
Stels [109]

Answer:

115.05 Volts

Explanation:

We can use Ohm's Law!

V=IR

V= 1.77 * 65

V=115.05 Volts

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3 years ago
THIS IS SOOO IMPORTANT PLS HELP I’LL GIVE BRAINIEST
Ket [755]

Answer:

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Explanation:

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6 0
3 years ago
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0
3 years ago
Read 2 more answers
A locomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (w
VMariaS [17]

Answer:

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

3 0
3 years ago
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