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DIA [1.3K]
2 years ago
8

A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is

the frictional force?
If work is shown that would be great
Physics
1 answer:
Blizzard [7]2 years ago
5 0

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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

##############################

If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

_________________________________

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0
3 years ago
A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.
kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

\displaystyle t=\sqrt{\frac{2h}{g}}

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

\displaystyle d=v.t

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

\displaystyle t=\sqrt{\frac{2*20}{9.8}}

\displaystyle t=\sqrt{4.0816}

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

\displaystyle d=10\cdot 2.02

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0
3 years ago
I need help with this
mojhsa [17]

Answer:

1) distance walked = 210 m

2) not sure

3) avg speed = total d/ total t = 210/52 = 4 m/s approx

4) For avg vel, use the formula - displacement/time

Explanation:

im not fully sure but i studied this last year

7 0
3 years ago
Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:
vodka [1.7K]

Answer:

nolo se

Explanation:

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8 0
3 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
8 0
3 years ago
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