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givi [52]
4 years ago
13

How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?

Physics
1 answer:
Andrej [43]4 years ago
6 0

Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

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Which device is used to measure interior diameter of a water pipe and in which units are the callibrations​
ahrayia [7]

We can use a vernier calliper

3 0
3 years ago
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A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi
S_A_V [24]

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

6 0
3 years ago
Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jekas [21]

Answer:

F_2 = 29.54 N

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

\tau = \vec r \times \vec F

here we will have

\vec r_1 \times F_1 = \vec r_2 \times F_2

so we have

13 \times 50 = 22 \times F_2

so we have

F_2 = \frac{13 \times 50}{22}

F_2 = 29.54 N

8 0
3 years ago
. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d
NikAS [45]

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

3 0
3 years ago
HELLPPP NOWWA!!!!!!!!!!!!
prisoha [69]

Answer:

b

Explanation:

the gravitational pull also helps with that but

7 0
3 years ago
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