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4 years ago

Because not all stars are the same distance from earth, it can be difficult to determine?

Physics

2 answers:

Inessa [10]4 years ago

8 0

I think answer is A. The Brightness of a Star

ehidna [41]4 years ago

8 0

The answer is b because some star are really bright which make us think that they are close but they are probably really far

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1. What is the efficiency of a heat engine that does 1250 J of work and gives off 5250 J of heat to

vladimir2022 [97]

answer: in pic

explanation: found it here http://gaaq.weebly.com/uploads/7/8/8/4/78844490/chapter_11_q_as.pdf

7 0

3 years ago

Read 2 more answers

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is 54.55 %.

Learn more here:

brainly.com/question/20793607

3 0

3 years ago

Identify the part of a carrot plant that contains chlorophyll and the part that contains

Alex Ar [27]

Answer:

Leaves and Roots

Explanation:

The leaves of carrots contain chlorophyll which gives them their green color.

Stored in the carrot roots is extra glucose that serves as a food source for the second year's growth of the carrot plant. The carrot plant needs this stored eneergy to reproduce in its second year.

8 0

3 years ago

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po

liberstina [14]

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/ $s^2$ . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

$v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s$

So jumping velocity of Julia = 8 m/s

6 0

3 years ago

As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.

Ganezh [65]

Answer: (a) The bicycle is ahead of the car for 4 s.

(b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

$v=u+at$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

$22.3=0+4.02t$

$t=\frac{22.3}{4.02}$

t= 5.5 s

Use the equation of the motion to calculate the time taken by the bicycle.

$v=u+at_{1}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

$8.94=0+5.81t_{1}$

$t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s$

Calculate the time interval for which the bicycle is ahead of the car.

$t-t_{1}= 5.5 s - 1.5s$

$t-t_{1}= 4s$

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

$S=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

$S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}$

$S= 60.8 m$

Use the equation motion to calculate the distance covered by the bicycle.

$S_{1}=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

$S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}$

$S_{1}= 5.8 m$

Calculate the maximum distance covered by the bicycle to lead the car.

$S-S_{1}=60.8-5.8=55m$

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0

3 years ago