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Citrus2011 [14]
3 years ago
5

A car whose mass is 1000kg is traveling at a constant speed of 10 m/s2. Neglecting any friction, how much force will the engine

have to supply to keep going the same speed?
Physics
1 answer:
matrenka [14]3 years ago
7 0
Without friction, NO FORCE is required to keep a moving object moving
at a constant speed in a straight line. 

Force CHANGES speed or direction.  If neither speed nor direction changes,
then there's no force acting on the object.
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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s^2. The friction experienced by the swimmer is 100 N. How many N
Helga [31]

Answer:

490N

Explanation:

According Newton's second law!

\sum Force = mass × acceleration

Fm - Ff = ma

Fm is the moving force

Ff s the frictional force = 100N

mass = 65kg

acceleration = 6m/s²

Required

Moving force Fm

Substitute the given force into thr expression and get Fm

Fm -100 = 65(6)

Fm -100 = 390

Fm = 390+100

Fm = 490N

Hence the force that will cause two cart to move is 490N

5 0
2 years ago
Zinc metal (Zn) and sulfur powder (s) undergo a chemical reaction to form zinc sulfide (ZnS) which equation represents this chem
dexar [7]

Zinc metal (Zn) reacts with sulfur (S) to create zinc sulfide (ZnS), and the chemical reaction is: Zn (s) + S (s) = ZnS (s).

Importance of Zinc (Zn) and sulfur (S):

  • Zinc(Zn) is a vital element that our systems need to absorb food and nutrients as well as create healthy skin and bones. Zinc(Zn) ions play a crucial role in a number of the body's enzymes.
  • Sulfur(S) is a pale yellow, tasteless, brittle solid that is also necessary for life. It is found in many proteins as well as the amino acids cysteine and methionine. It is a trace element found in bone minerals, bodily fluids, and lipids.

Chemical reaction -

In this experiment, heating a zinc(Zn) and sulfur(S) combination causes an interesting chemical reaction. A blinding flash of light, hot sparks, a hissing sound, and a cloud of white smoke in the shape of a mushroom are produced.

Therefore, the following chemical processes are taking place in the reaction: Zn (s) + S (s) = ZnS (s).

Learn more about Zinc(Zn) here:

brainly.com/question/13890062

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8 0
2 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
3 years ago
Read 2 more answers
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Umnica [9.8K]

Answer:

d and i think a?

Explanation:

im so sorry if its wrong

5 0
3 years ago
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