All categories

2 years ago

Which example is correctly matched with its type of friction?

Physics

1 answer:

vlabodo [156]2 years ago

3 0

Answer:

A. pushing a car that isn't moving

You might be interested in

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

RideAnS [48]

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

6 0

2 years ago

Read 2 more answers

What must happen to the temperature of a material for thermal expansion to occur.

Aloiza [94]

In most cases the temperature must increase for thermal expansion to occur. Most substances expand as temperature increases because the atoms or molecules vibrate faster as temperature increases and experience greater separation.

8 0

2 years ago

) An ideal Carnot engine extracts 529 J of heat from a high-temperature reservoir during each cycle, and rejects of heat to a lo

enot [183]

Answer:

0.5747 or 57.4%

Explanation:

Your question is not complete so i will assume the right question goes thus;

A Carnot engine extracts 529 J of heat from a high-temperature reservoir during each cycle, and rejects 225 J of heat to a low-temperature reservoir during the same cycle. What is the efficiency of the engine?

Efficiency of a Carnot engine is given by [(Th-Tc) / Th]

using the formula above, the resulting equation will give us; (529-225)/529

=0.574669

3 0

2 years ago

Omar wrote a hypothesis about batteries called dry cells. If more dry cells are connected end to end, a light bulb will work lon

iren2701 [21]

Answer:

The independent variable is the number of dry cells and the dependent variable is the time the bulb works.

Explanation:

In this exercise, you are asked to analyze the variables derived from Ómar's hypotypeis

"If more dry cells are connected end-to-end, a light bulb will work longer because more energy is available."

In this hypothesis, the independent variable that is controlled by the researcher is the number of batteries to be connected in series.

The dependent variable that is measured by the researcher is how long the bulbs last.

When reviewing the different answers, the correct one is:

The independent variable is the number of dry cells and the dependent variable is the time the bulb works.

7 0

2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

Speed of blue car after 2.6 seconds of starting the motion:

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

Speed of blue car after 7.2 seconds of starting the motion:

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

Distance travelled by the blue car before application of brakes:

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

Now the distance travelled during the accelerated motion:

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

Now the distance travelled at uniform motion:

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

Acceleration of the blue car once the brakes are applied

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

The total time for which the blue car moves:

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

Now the time taken to stop the blue car after application of brakes:

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

For the acceleration of the yellow car:

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

2 years ago