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damaskus [11]
3 years ago
12

A snail con move at 3cm/s if it’s ke is 300joules what it it’s mass?

Physics
1 answer:
skelet666 [1.2K]3 years ago
3 0

Explanation:

V = 3 cm/s = 0.03 m/s. BY THE FORmULA OF K.E. K.E = 1/2 mV^2. 300 =1/2 m (0.03)^2. m = 300 x 2/0.0009.

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Emilio pushes a 100 kg freshman with 200 N of force. How much is the freshman accelerated?
ladessa [460]

Explanation:

F = MA

200 = 100 * A

A = 200/100

A = 2m/sec^2

<h3><em>hope </em><em>it </em><em>helps </em><em>you </em></h3>
8 0
3 years ago
A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?
olchik [2.2K]
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
5 0
4 years ago
What are the two kinds of nucleons?
Akimi4 [234]

Answer:

The main types of nucleons are protons and neutrons. A proton, as its name suggests, has a positive electric charge, and a neutron has a neutral electric charge (meaning that it has no charge). The two in the nucleus of the atom make a positive charge, since the neutron has no charge at all.

Explanation:

5 0
3 years ago
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
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