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ANEK [815]
3 years ago
8

A car is driving directly north on the freeway at a speed of 91 km/h and a truck is leaving the freeway driving 56 km/h in a dir

ection that is 28° west of north. What is the velocity of the truck relative to the car? Please report the answer with a positive angle.
Physics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

v = 49.69 km/hr

Explanation:

velcoity of truck with respect to car is given as = velocity of truck - velocity of car

velcoity of truck with respect to car is given as = (-54 sin(28)i +54cos(28)j - 91km/h

                                                                              = -24.35i - 43.32j

magnitude of velocity is

v = \sqrt{ 24.35^2+43.32^2}

v = 49.69 km/hr

the direction is

tan^{-}[\frac{ 43.32}{24.35}]

= 60.65 degree towrd south of west

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What is the frequency of a wave with a wavelength of 12 meters and a velocity of 4 m/s?
trasher [3.6K]

Answer:

0.33 hz

Explanation:

the formula for the frequency in this situation is f=v/wavelength

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3 years ago
A bag of sugar weighs 2 kg on earth. What should it weigh in newtons on the moon, where the free-fall acceleration is 1/6 that o
Elis [28]
<span> Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
1 kg = 2.2 lbs, so 2.0 lbs x 1 kg/2.2 lbs = 0.91 kg
The bag would have a weight of 9.8 x 0.91 = 8.9 N

1. 8.9 x 1/6 = 1.5 N

2. 8.9 x 2.64 = 23.5 N

The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight. </span>
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The half-life of carbon-14 is 5730 years. how long will it take for 7/8 of a sample of carbon-14 to decay
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5 0
3 years ago
Read 2 more answers
How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
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