1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
konstantin123 [22]
3 years ago
8

A small car is traveling at a speed of 60 mph on the highway. In the next lane, a large passenger bus is traveling at the same s

peed. Which of the following is true?
A. The bus has more kinetic energy than the car.
B. The car has more kinetic energy than bus.
C. The car and the bus have potential energy, not kinetic energy.
D. The car and bus have the same amount of kinetic energy.
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

A. The bus has more kinetic energy than the car.

Explanation:

Kinetic energy is the amount energy possessed by the body when in motion. Kinetic is given by the formula;

K.E = 1/2mv²,

where m is the mass of the body and v is the velocity of the body.

Kinetic energy therefore depends on the mass and the velocity of the body. In this case, the passenger bus has more kinetic energy at constant speed than the car since it has a larger mass.

You might be interested in
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
1. What is the formula for the period of a pendulum and what is the main determining factor in its period?
Brut [27]

Answer:

T=2\pi \sqrt{\frac{L}{g}}

Explanation:

A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.

The period of a pendulum is the time it takes for the pendulum to complete one oscillation.

The period of a pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that the period of a pendulum does not depend on the mass.

Therefore, the only 2 factors affecting the period of a pendulum are:

- The length of the pendulum: the longer it is, the longer the period of oscillation

- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum

7 0
3 years ago
WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!
LenKa [72]

Answer:Weight is the result of gravity . The gravitational field strength of Earth is 10 N/kg (ten newtons per kilogram). This means an object with a mass of 1 kg would be attracted towards the centre of Earth by a force of 10 N. We feel forces like this as weight.

Explanation:

3 0
3 years ago
Read 2 more answers
Which factor would reduce the size of a forest population
zhuklara [117]
C. Lack of mates. If they cannot reproduce enough, their size will reduce.
7 0
3 years ago
Read 2 more answers
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0
3 years ago
Other questions:
  • Is winter eye grass autotrophic
    15·2 answers
  • The property that describes how radily a substance combines chemically with other substances is _____________
    11·1 answer
  • Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
    15·1 answer
  • how long will it take an airplane yo go 800 miles if its traveling at an average speed of 650 miles per hour
    6·1 answer
  • What are chemical properties
    11·2 answers
  • A voltage amplifier with an input resistance of 40k ohms, an output resistance of 100 ohms, and a gain of 300 V/V is connected b
    6·1 answer
  • Why do automobiles rust faster in wet climates than in drier climates?
    6·2 answers
  • A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the
    5·1 answer
  • Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula
    8·1 answer
  • PLZZZ HELP!!!<br><br><br> State the forces in a spring, magnets, electric scales, and a newton metre
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!