While John is traveling along a straight interstate

What percentage of the earth's mass is the moon's mass?

Schach [20]

Answer:

ssssssssss

Explanation:

hshjdjjfjjjjhhddd

5 0

3 years ago

You are out running and the first mile takes you 10 minutes. The second mile takes you 20 minutes. This is an example of

cupoosta [38]

If Equal distance is traveled in equal interval of time then it is known as uniform motion in which velocity of object will remain same.

Then if distance covered will be same and the time taken to cover same distance is decreasing then it shows that speed is increasing with time due to which it took less time to cover same distance. This is also known as positive acceleration.

Now if the distance covered will be same and time taken to cover same distance is increasing then it shows that speed is decreasing with time due to which it took more time to cover the same distance. This is also known as negative acceleration.

Now in the above case it is given that the first mile takes you 10 minutes. The second mile takes you 20 minutes. So the time taken is increasing while we cover same distance so this is an example of <u>Negative Acceleration</u>

5 0

3 years ago

Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o

adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

$F = \sqrt{F_h^2+F_v^2}$

$F = \sqrt{3.36^2+13.98^2}$

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

$\theta =tan^{-1}(\dfrac{F_v}{F_h})$

$\theta =tan^{-1}(\dfrac{13.98}{3.36})$

θ = 76.48°

3 0

3 years ago

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N