All categories

3 years ago

If a car travels around a gentle curve on a highway at 60 km/h, does the velocity change

1 answer:

4 0

Yes the velocity changes. Because velocity changes with direction. The object is moving around a gentle curve. The curve is not linear it is curve the direction changes a bit so obviously the velocity also changes but not much. Juts a minor change. Depends on how much curve the highway is.

You might be interested in

Find the change in kinetic energy of a 0.650 kg fish leaping to the right at 15.0 m/s that collides inelastically with a 0.950 k

yaroslaw [1]

The change in the kinetic energy refers to the work done in displacing a body, thus, the change in the kinetic energy of an object refers to the work done on the object.
The correct formula to use is:
W = Initial kinetic energy - Final kinetic energy;
Where, W = change in kinetic energy
Final kinetic energy and initial kinetic energy = 1/2 MV^2
Initial velocity = 15 m/s
Final velocity = 13.5 m/s
Initial mass = 0.650 kg
Final mass = 0.950 kg
W = 1/2 [0.650* (15 *15)] - 1/2 [0.950 * (13.5 * 13.5)]
W = 146.25 - 173.13 = 26.88
Therefore, the change in kinetic energy is 26.88 J.
The negative sign has to be ignored, because change in kinetic energy can not be negative.

4 0

3 years ago

Read 2 more answers

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

Debora [2.8K]

You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.

7 0

3 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

If a wave has amplitude of 2 meters, a wavelength of 2 meters, and a frequency of 10 Hz, and a period of 1 second, then at what

Serhud [2]

Answer:

20 m/s

Explanation:

The speed of a wave is given by:

$v=\lambda f$

where

$\lambda$ is the wavelength

f is the frequency

v is the speed

For the wave in this problem,

f = 10 Hz is the frequency

$\lambda=2 m$ is the wavelength

So the speed is

$v=(10 Hz)(2 m)=20 m/s$

6 0

3 years ago

Why is a flower not a good blackbody radiator?

artcher [175]

Answer:

The answer is O C. A flower absorbs most of the light that hits it.

Explanation:

blackbody radiator is defined as an object that absorbs all electromagnetic radiation that falls on it at all frequencies over all angles of incidence.
No radiation is reflected from such an object. According to thermodynamic arguments embodied in Kirchhoff's law, a good absorber is also a good emitter.

3 0

1 year ago