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Rufina [12.5K]
3 years ago
10

A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the

push?
1,500 W
5,000 W
112 W
450 W
Physics
2 answers:
Tamiku [17]3 years ago
6 0

The work done to push the puck is the product between the force F and the distance d:

W=Fd=(750 N)(2.0 m)=1500 J


And the power is the work done (W) per unit time (t):

P=\frac{W}{t}


Substituting the work done, W=1500 J, and the time taken, t=0.30 s, we find the power of the push:

P=\frac{1500 J}{0.30 s}=5000 W



lubasha [3.4K]3 years ago
6 0
Power, P = Force, F * Velocity, V

V = distance/time = 2/0.3 = 6.67 m/s

Therefore,
P = 750*6.67 = 5000 W

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Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7
Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

6 0
3 years ago
A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?
poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

5 0
3 years ago
A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m
solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0
3 years ago
A plane flying against a jet stream will travel faster than a plane traveling with a jet stream. Please select the best answer f
stepan [7]

"with the wind" is a tail-wind, and the speeds are added to get the groundspeed.

"against the wind" is a head-wind, and the windspeed is subtracted from the airspeed.

3 0
3 years ago
Read 2 more answers
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
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