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3 years ago

Which of the following questions best highlights a shortcoming of the big bang theory?

Physics

2 answers:

kaheart [24]3 years ago

4 0

<span>A. How could energy become the matter present today? </span>

quester [9]3 years ago

4 0

Answer:

What was the cause of the big bang theory?

Explanation:

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A refrigerator is used to remove 84 kj/min of heat from a tank. If the electric power consumed by the refrigerator is 1. 2 kw, w

zvonat [6]

Answer:

84 kj/min = 1.4 kj/sec

Power Out / Power In = Heat Out / Heat In - Coefficient of Performance

1.4 kj/sec / 1.2 kj/sec = 1.17 = COP

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2 years ago

During a football game, the running back changes direction from running 4.5 m/s [N] to 4.5 m/s [S] over 8 s. What is the acceler

Nimfa-mama [501]

Explanation:

Take north to be positive and south to be negative.

a = (v − v₀) / t

a = (-4.5 m/s − 4.5 m/s) / 8 s

a = -1.125 m/s²

The acceleration is 1.125 m/s² south.

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2 years ago

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:

- A uniform motion with constant velocity along the horizontal direction

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

$u_y=0$ is the initial vertical velocity (the missile is thrown horizontally)

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s$

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

$v_x = 45 m/s$

Therefore, the distance covered in a time t is

$d=v_x t$

and by substituting t = 3.91 s, we find the horizontal range of the missile:

$d=(45)(3.91)=176 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago

A sled of mass 80 kg starts from rest and slides down an 18° incline 90 m long. It then travels for 20 m horizontally before sta

yaroslaw [1]

Answer:

net work done by friction = - 2205 J

Explanation:

given,

mass of sled = 80 kg

slide down at an angle of 18°

length = 90 m

travel horizontally = 20 m before starting back up a 16° incline.

net work = ?

$h_i = initial\ height$

$h_i = L_i Sin 18^0$

$h_i = 90 sin 18^0$

$h_i = 27.81\ m$

$h_f = Final\ height$

$h_f = L_f Sin 16^0$

$h_f = 90 sin 16^0$

$h_f =24.81\ m$

net work done by friction = change in potential energy

= mg (h_f - h_i)

= (75) (9.8) (- 27.81 + 24.81)

net work done by friction = - 2205 J

7 0

3 years ago

A stone whirled at the end of the a rope 30cm long, makes 10 complete resolution in 2 seconds Find: A. The angular velocity in r

lord [1]

Answers:

A: Angular velocity $\omega=31.40 \frac{r a d}{s}$

B: Linear velocity $v=9.42 \frac{m}{s}$

C: Linear Distance $d=47.1 \mathrm{m}$

Given:

Radius of the rope r=30cm=0.3m

Angular distance $\Delta \theta$ =10 revolutions

Time taken t=2seconds

To find:

A: Angular velocity in radians

B: Linear speed

C: Distance covered in 5 seconds

<u>Step by Step Explanations:</u>

Solution:

A: Angular velocity in radians;

According to the formula, Angular velocity can be calculated as

Angular Velocity = angular distance/ time

$\omega=\Delta \theta / \Delta t$

Where $\omega$ =Angular velocity

$\Delta \theta$ =Angular distance=10 revolutions

Changing revolutions to radians multiply with $2 \pi$ , so that we get

$=10 \times 2 \pi$

$=10 \times 2(3.14)$

=62.80 rad/rev

$\Delta t$ =Change in time

Substitute the known values in the above equation we get

$\omega$ =62.80 / 2

$\omega=31.40 \frac{r a d}{s}$

B. Linear speed of the rope;

As per the formula

Linear speed = angular speed × radius

$v=\omega \times r$

Where $\omega$ =Angular velocity

v=Linear speed of the rope

r=Radius of the rope

Substitute the known values in the above equation we get

$v=31.40 \times 0.30$

$v=9.42 \frac{m}{s}$

C. Dsitance covered in 5 seconds;

Linear distance = linear speed × time

$d=v \times t$

Where d= Linear distance of the rope

v=Linear speed of the rope

t=Time taken

Substitute the known values in the above equation we get

$d=9.42 \times 5$

$d=47.1 \mathrm{m}$

Result:

Thus A: Angular velocity of the rope $\omega=31.40 \frac{r a d}{s}$

B Linear speed of the rope $v=9.42 \frac{m}{s}$

C: Distance covered in 5 seconds $d=47.1 \mathrm{m}$

6 0

3 years ago