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Arturiano [62]
3 years ago
12

A person takes a trip, driving with a constant speed of 94.5 km/h except for a 22.0 min rest stop. If the person's average speed

is 65.1 km/h, how much time is spent on the trip and how far does the person travel?
Physics
1 answer:
qwelly [4]3 years ago
4 0
From the average speed you can fix an equation:

Average speed = distance / time

You know the average speed = 65.1 kg / h, then

65.1 = distance / total time,

where total time is the time traveling plus 22.0 minutes

Call t the time treavelling and pass 22 minutes to hours:

65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1

 
From the constant speed, you can fix a second equation

Constant speed = distance / time traveling

94.5 = distance / t ==> distance = 94.5 * t

The distance is the same in both equations, then you have:

[t +22/60] * 65.1 = 94.5 t

Now you can solve for t.

65.1t + 22*65.1/60 = 94.5t

94.5t - 65.1t = 22*65.1/60

29.4t = 23.87

t = 23.87 / 29.4

t  = 0.812 hours

distance = 94.5 km/h * 0.812 h = 76.7 km

Answers: 1) 0.81 hours, 2) 76.7 km


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Anastasy [175]

Answer:

The object will travel at the speed of 16 m/s.

Explanation:

Given

  • Mass m = 4.0 kg
  • Momentum p = 64 kgm/s

To determine

How fast is the object traveling?

<u>Important Tip:</u>

The product of the mass and velocity of an object —  momentum.

Using the formula

p = mv

where

  • m = mass
  • v = velcity
  • p = momentum

Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula

p = mv

64\:=\:4\times v

switch the equation

\:4\times \:v\:=64

divide both sides by 4

\frac{4v}{4}=\frac{64}{4}

simplify

v=16 m/s

Therefore, the object will travel at the speed of 16 m/s.

3 0
3 years ago
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the
Over [174]

Answer:

Part a)

a= 0.32 m/s^2

Part b)

F_c = 3.6 N

Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

F_f = 3.45 N

Now we know by Newton's II law

F_{net} = ma

so we have

F_p - F_f = (m_a + m_b) a

9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

a= 0.32 m/s^2

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

3 0
3 years ago
What is the mechanical advantage of the wheel and axle shown below?
BigorU [14]

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and  r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

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6 0
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exis [7]

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

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Luda [366]

Answer:

0.0334N

Explanation:

Given parameters:

M1  = 5 x 10⁶kg

M2  = 1 x 10⁶kg

Distance  = 100m

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

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G is the universal gravitation constant

m is the mass

r is the distance

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