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liq [111]
3 years ago
7

Why can Muhammad exert a greater punching force with his bare fist than he can while wearing a

Physics
1 answer:
weqwewe [10]3 years ago
5 0

Answer:

the glove could be heavy so slowing down his power

Explanation:

JUST GUESSED

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The 94-lb force P is applied to the 220-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Makovka662 [10]

Answer

given,

force = 94 lb

weight of crate = 220 lb

Assuming the static friction be equal = 0.47

                       kinetic friction = 0.36

Maximum force applied to move the object is when object is just start to move.

F = μ N

F = 0.47 x 220

F = 103.4 lb

As the frictional force is more than applied then the object will not move.

so, the friction force will be equal to the force applied on the object that is equal to 94 lb.

hence, the direction of force will left.

8 0
3 years ago
When searching for your word processing file to finish writing your report, you should look for a file with which extension?
ryzh [129]

Answer:

doc

Explanation:

6 0
2 years ago
Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5
lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0
2 years ago
Who was the english thinker who established the three laws of motion?.
Tatiana [17]

Answer:

Isaac Newton

Explanation:

Because i learned this in school

4 0
2 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
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