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Alex73 [517]
3 years ago
15

What is the velocity of a group of explorers traveling from Antarctica to Australia, traveling 1500 miles in 48 hours?31.25 mph

south
72,000 mph
48 mph
31.3 mph north
Physics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

31.3mph North

Explanation:

Given parameters:

Distance traveled  = 1500miles

Time taken  = 48hrs

Unknown:

Velocity of the group of explorers = ?

Solution:

Velocity is the displacement divided by the time taken. It is a vector quantity with both magnitude and direction.

 So;

       Velocity = \frac{displacement}{time}

Displacement is the distance moved in a specific direction = 1500miles North

 Insert the parameters and solve;

   Velocity = \frac{1500}{48}   = 31.3mph North

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Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

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  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
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3 years ago
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
2 years ago
Amy runs 2 blocks south, then turns around and runs 3 blocks north. what is the distance and displacement
Elenna [48]

Answer: Distance=  5 blocks and displacement= -1 block

Explanation: distance is the total length of a journey while displacement is the shortest straight length of a journey!

5 0
3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

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