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xenn [34]
3 years ago
13

If a projectile fired beneath the water, straight up, breaks through the surface at a speed of 13m/s, to what height above the w

ater will it ascend
Physics
1 answer:
OlgaM077 [116]3 years ago
3 0
Well if you had either the velocity or distance traveled i could tell you. But since you haven't all i can say for sure is that the water slowed the bullet down to 13m/s so lets say you knew the distance you would calculate how many meters it traveled and you would have your answer because in this situation, meters (height) =how many seconds spent going into the air. 
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Two objects of masses m1 = 0.56 kg and m2 = 0.88 kg are placed on a horizontal
ziro4ka [17]

Explanation:

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3 0
2 years ago
A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s
marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0
3 years ago
An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
3 years ago
A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

6 0
3 years ago
If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats
nekit [7.7K]

Your answer should be 4Hz

8 0
3 years ago
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