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Inessa [10]
3 years ago
15

You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you

r base camp, after walking for 5 h?
Physics
1 answer:
Leona [35]3 years ago
4 0
Http://www.calculator.net/pace-calculator.html?ctype=distance&ctime=05%3A00%3A00&cdistance=5&cdistanceunit=Miles&cpace=02%3A00%3A00&cpaceunit=tpm&printit=0&x=87&y=24 a pace calculator
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Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107m above Earth's surface (that is, if it were three Earth ra
Vladimir79 [104]
The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N
5 0
3 years ago
A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.​
mars1129 [50]

Answer:

18,316.2

Explanation:

The formula for GPE is mgh, where

M = Mass of the object

G = Acceleration due to gravity (9.8 m/s^2 on Earth)

H = Height above ground

890 × 9.8 × 2.1 = 18,316.2

4 0
3 years ago
Read 2 more answers
The type of energy that depends on position is called
Maurinko [17]

Answer:

a. potential energy

Explanation:

ur welcome

3 0
3 years ago
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A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj
Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area  

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = \sigma_B= \frac{M_y}{I} where I = \frac{(0.5^2 + 4^2)}{12}

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0
3 years ago
Read 2 more answers
Why does a person while firing a bullet holds the gun tightly to his shoulder? SHORT ANSWER ​
Amanda [17]

Answer:

They hold it tightly to their shoulder because the gun recoils and jumps back so by holding it to their shoulder it allows them to stablize the gun so that when it jumps in stead of the gun kicking back and flying the gun jumps and hit the shoulder.

Explanation:

6 0
3 years ago
Read 2 more answers
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