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2 years ago

12

Peter left town A at 13:30 and travelled towards town B at an average speed of 40mph. At 13:45 Philip left town A for town B at

an average speed of 30 mph. What was the distance between them at 15:00?

1 answer:

skad [1K]2 years ago

8 0

Answer:

fbddb

Explanation:

dsbse

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Ms. Mary Mack walked around her block from her house for 200 meters. She arrived back at her house in 15 minutes. What was her d

inysia [295]

Answer:

d = 0 [m]

Explanation:

Displacement is understood as the length and direction that a body travels to move from an initial point to an endpoint.

This displacement is represented with a vector or straight line that indicates the distance of the displacement and its length.

This displacement in an easier way to understand. It is the distance between the start point and the endpoint of the journey. Since the second point is equal to the first point, since Mary returns to the same place, there is no difference between the displacement.

Therefore the displacement is zero.

5 0

3 years ago

a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0

2 years ago

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

$T=\frac{2\pi r}{v}$

Re-arranging for r

$r=\frac{Tv}{2\pi}$

And substituting into the previous equation

$qvB = m \frac{Tv^3}{2\pi}$

Solving for T,

$T=\frac{2\pi q B}{m v^2}$

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0

3 years ago

Protons and neutrons grouped in a specific pattern

alexgriva [62]

Answer b protons and electrons

5 0

3 years ago

In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?

IceJOKER [234]

Answer: $a=1016.66 m/s^{2}$

Explanation:

Acceleration $a$ is expressed in the following formula:

$a=\frac{V_{f}-V_{o}}{t}$

Where:

$V_{f}=610 m/s$ is the final velocity of the projectile

$V_{o}=0 m/s$ is the initial velocity of the projectile

$t=0.6 s$ is the time

Solving:

$a=\frac{610 m/s-0 m/s}{0.6 s}$

$a=1016.66 m/s^{2}$ This is the acceleration of the projectile

6 0

3 years ago