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polet [3.4K]
3 years ago
5

URGENT!! HELP

Physics
2 answers:
Monica [59]3 years ago
8 0

ill try to get back to you, im doing the test now

Zina [86]3 years ago
7 0

Answer:

Did you get your answer for Number 15?

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A 70.0 kg ice hockey goalie, originally at rest, has a 0.170 kg hockey puck slapped at him at a velocity of 33.5 m / s . Suppose
Sergeeva-Olga [200]

Answer:

Final velocity of both goalie & puck = 0.018116 m/s

Explanation:

M1U1 + M2U2 = (M1+M2)V

70 x 0 + 0.17 x 33.5 = (70+0.17)V

V = 0.08116m/s

4 0
4 years ago
A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with t
steposvetlana [31]

Answer:

0.35

Explanation:

According to Newton's second law;

\sum Fx = ma

Fm - Ff =ma

Fm is the moving force = Wsin theta

Fm = 4(9.8)sin55

Fm = 32.1N

Ff is the frictional force = nmgcos theta

Ff = n(4)(9.8)cos55

Ff = 22.48n

Acceleration a = 6.0m/s²

Substitute the given values into the formula and get the coefficient of friction

32.11-23.48n = 4(6)

32.11-24= 23.48n

8.11 = 23.48

n = 8.11/23.48

n = 0.35

Hence the coefficient of friction is 0.35

6 0
3 years ago
Which of these is a benefit of social networking ?
Alisiya [41]

Answer:

Staying connected to friends

Explanation:

hope this helps

6 0
3 years ago
GIVING BRAINLIST!!!!
34kurt
A negative object so it takes in the charged object and nothing will happen until something else would touch it I think
3 0
3 years ago
Read 2 more answers
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

7 0
3 years ago
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