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polet [3.4K]
3 years ago
12

A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

ion is 0.25. What is the force of friction on the box? What is the acceleration of the box?
Physics
2 answers:
fgiga [73]3 years ago
5 0

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

Burka [1]3 years ago
4 0
Yeah what that guy said lol
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Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro
luda_lava [24]

Answer:

c. 4.582\times10^{21} kg

Explanation:

r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

v_{i} = Initial speed of rock = 136 ms⁻¹

v_{f} = Final speed of rock = 392 ms⁻¹

m = mass of the rock

M = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

8 0
3 years ago
Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with
ad-work [718]

Answer:

y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

         y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

         0 = y₀ + v₀ t - ½ g t²

        y₀i = -v₀ t + ½ g t²

let's calculate

         y₀ = - 8  2.5 + ½  9.8  2.5²

         y₀ = 10.625 m

7 0
3 years ago
Who is known as the "father of modern chemistry" because he first organized all known elements into four different
iVinArrow [24]

Answer:

D. Henry Moseley

Explanation:

I'm chemistry right now

5 0
3 years ago
Read 2 more answers
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
The fuzzy cloud around the nucleus is called the
Kamila [148]
It's number three on your worksheet. ;)
6 0
3 years ago
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