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3 years ago

Convert 6.6 meters to centimeters

Physics

2 answers:

Paraphin [41]3 years ago

7 0

680 cm

hope this helps

Lina20 [59]3 years ago

6 0

Answer:

660 centimeters

Explanation:

There are 100 cm in 1 m. To convert from m to cm, multiply by 100.

$6.6 \times 100 = 660$

There are 660 cm in 1 m.

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What happens to the heat energy when you increase the length of an object

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Answer: When the temperature of an object increases, the average kinetic energy of its particles increases. When the average kinetic energy of its particles increases, the object's thermal energy increases. Therefore, the thermal energy of an object increases as its temperature increases.

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3 years ago

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex

ratelena [41]

I am pretty sure that when you travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an example of negative acceleration. I consider this to be correct because <span>the second mile was slower. Hope you will agree with me. Regards!</span>

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3 years ago

Read 2 more answers

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

A woman exerts a constant horizontal force on a large box. As a result, the box moves across a horizontal floor at a constant sp

d1i1m1o1n [39]

Answer:C

Explanation:

When a constant horizontal force is applied to the box, box started moving in the horizontal direction such that it moves with constant velocity $v_0$

Constant velocity implies that net force on the box is zero

i.e. there must be an opposing force which is equal to the applied force and friction force can serve that purpose.

So option c is the correct choice.

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3 years ago

Whais the cause of necular bomb

coldgirl [10]

Answer:

The atomic bomb, is as defined by britannica.com “a deadly weapon caused by the sudden release of energy after the splitting, or fission, of the nuclei of heavy elements like uranium.” In 1945, the United States (US) dropped two atomic bombs, one in Hiroshima and the other in Nagasaki ending WWII.

Explanation:

Hope this helps :)

pls mark brainliest :P

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2 years ago