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Phantasy [73]
3 years ago
5

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

ring

Physics
2 answers:
AnnZ [28]3 years ago
5 0
Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

Sedaia [141]3 years ago
4 0

<u>Answer:</u> The tension in the string is 6 N

<u>Explanation:</u>

Tension is defined as the pulling force that is applied on a string, chain etc.. Force is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=ma

where,

F = tension or force exerted in the string

m = mass of the cart pulling the string = 3.00 kg

a = acceleration of the string = 2.00m/s^2

Putting values in above equation, we get:

F=3.00kg\times 2.00m/s^2\\\\F=6N

Hence, the tension in the string is 6 N

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a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work
ioda

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

a=g=9.8 m/s^2 is the acceleration of gravity

Solving the equation for v, we find the final velocity:

v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s

So, the final speed is 12.5 m/s.

3 0
3 years ago
A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit
barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

4 0
3 years ago
How did harlow shapley conclude that the sun was not in the center of the milky way galaxy?
timurjin [86]

We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.

<h3>What is the work of Harlow Shapley?</h3>

Shapley, who was headquartered in Boulder, Colorado, used Cepheid variable stars to estimate the size of the Milky Way Galaxy and its position relative to the Sun. In 1953, he published his "liquid water belt" theory, today known as the concept of a livable zone.

There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.

Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.

To learn more about Harlow Shapley's original estimate go to - brainly.com/question/28145909

#SPJ4

3 0
2 years ago
An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

5 0
3 years ago
Which element is found in all the acids
lana66690 [7]
Hydrogen .When acids touches all metal hydrogen gas is emitted .Strong acids is one that can produce a high concentration of hydrogen ions.Hope this helped!
3 0
3 years ago
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