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sp2606 [1]
3 years ago
9

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

ve in pulling the sled horizontally along the ground?
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0
If we consider that the surface wherein this box is located is frictionless in such a way that the friction does not affect its motion, we can calculate for the horizontal force by multiplying the force with the sine of the angle.
                     x-component of force = (4 N) x (sin 62°)
                                         = 3.53 N
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What is the slope of (0,0)(4,200)
Kruka [31]

Answer:

.02

Explanation:

X-x/ Y-y= 4-0/200-0 =.02

5 0
3 years ago
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Every year _______ species go extinct.<br> a. 3<br> b. 72<br> c. 1000<br> d. 20000
Tems11 [23]
Well, the thing is: we don't really know, as we don't even know how many species there are on earth.

If we take a look at the estimates of <span>World Wide Fund for Nature, an organization that works toward combating species extinction, their estimates vary from 200 to 100 000 - but a probable number is 20 000 (d). </span>
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7 0
3 years ago
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1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force
Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

4 0
4 years ago
Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of
sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass  2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁  = I₂ ω₂  ,  ω₁ and   ω₂ are angular velocities initially and finally .

I₁ / I₂     =  ω₂ / ω₁

I₁ / I₂     =  T₁ / T₂  , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3  x 2.3 x 10¹⁹ / M

= 1 + 5 / 3  x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0
3 years ago
What single property was the most important in jesseca's material
jek_recluse [69]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation: The Graphite was the material in the passage that had reflected most of the light.

6 0
3 years ago
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