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Montano1993 [528]
3 years ago
8

A sled is moving at a constant speed down a surface inclined at 45.0° with the horizontal and travels 30 meters in 4 seconds. Ca

lculate the vertical velocity of the sled at the end of 4 seconds.
Physics
1 answer:
slamgirl [31]3 years ago
5 0
To obtain the vertical component of the velocity, it is necessary to multiply the velocity at this time by sin 45 degrees. Since the velocity is not given, it must be solved first by dividing the distance by the time. To show this:

Velocity = distance / time
Velocity = 30 m / 4s
Velocity = 7.5 m/s

Vertical velocity = 7.5 sin 45
Vertical velocity = 5.3 m/s
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A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the
Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check:     V0 = 5 m/s

                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s     (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s  the speed after 300 sec

3 0
2 years ago
An object of mass 20kg is released from a height of 10 meters above the ground level. The kinetic energy of the the object just
Yuri [45]

Answer:

The answer is the object weighs 5 so 3-5 is 2

Explanation:

I took this test

7 0
2 years ago
Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
sergij07 [2.7K]

Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

  • charge on the alpha particle, q=2e=3.2\times 10^{-19}\ C
  • mass of the alpha particle, m=6.64\times 10^{-27}\ kg
  • strength of a uniform magnetic field, B=0.5\ T
  • radius of the final orbit, r=0.5\ m

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

m.v=q.B.r

where:

v = velocity of the alpha particle

v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

v=1.2048\times 10^{7}\ m.s^{-1}

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

6 0
3 years ago
The symbol used for magnetic field is:
alina1380 [7]
The letter "B" is used for magnetic field
5 0
3 years ago
The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex
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(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam
7 0
3 years ago
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