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labwork [276]
3 years ago
13

A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delive

red by the engine? (1 hp 746 W)
Physics
1 answer:
NARA [144]3 years ago
4 0

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

P = 67,000 W

P = 90 hp

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A boat is rowed perpendicular to the shore of a river that flows at 3.0 m/s as shown in the
vazorg [7]
The magnitude is 5 ms^-1

6 0
3 years ago
In separate experiments, a large number of particles (all with the same charge, but with a wide variety of masses and speeds) ar
likoan [24]

Answer:

  they have the same mass

Explanation:

The force applied by the field is a function of the charge and velocity, so the acceleration experienced by a particle will be dependent upon its mass. Particles in orbits with the same radius are exhibiting the same acceleration, so must have the same mass.

7 0
3 years ago
You repeated a measurement 6 times and recorded the time using a stop watch: 5.8s, 4.6s, 4.8s, 5.1s, 4.3s, 4.6s. What average va
iVinArrow [24]

Answer and Explanation

Arranging the measured values in increasing order;

4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s

The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.

So, the mean is the average of the four remaining data points after removing the outliers.

Mean = (4.6 + 4.6 + 4.8 + 5.1)/4

Mean = 4.775s

So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.

QED!

8 0
4 years ago
A 65 Kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. How fast will she be going at th
yan [13]

Answer:

34 m/s

Explanation:

Potential energy at top = kinetic energy at bottom + work done by friction

PE = KE + W

mgh = ½ mv² + Fd

mg (d sin θ) = ½ mv² + Fd

Solving for v:

½ mv² = mg (d sin θ) − Fd

mv² = 2mg (d sin θ) − 2Fd

v² = 2g (d sin θ) − 2Fd/m

v = √(2g (d sin θ) − 2Fd/m)

Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))

v = 33.9 m/s

Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.

8 0
3 years ago
A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do
Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0
3 years ago
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