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vodomira [7]
4 years ago
14

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

t their periods?
Physics
1 answer:
Elanso [62]4 years ago
6 0
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
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A 6.25-kg bowling ball moving at 8.95 m/s collides with a 0.95-kg bowling pin, which is scattered at an angle of θ = 27.5° from
Archy [21]

Answer:

-6.34 degrees.

Explanation:

Because this is a collision we can use the formula of linear momentum conservation.

m_1*v_{o1}+m_2*v_{o2}=m_1*v_{f1}+m_2*v_{f2}

The pin has an angle so it has two components of velocity, so:

v_x=V*cos(\theta)\hat{i}\\v_y=V*sin(\theta)\hat{j}\\\\v_x=11.6*cos(27.5^o)=(10.3m/s)\hat{i}\\v_y=11.6*sin(27.5^o)=(5.4m/s)\hat{j}

applying the first formula:

6.25kg*(8.95m/s)\hat{i}+0=6.25kg*v_{f}+0.95kg*(10.3m/s(\hat i)+5.4m/s(\hat j))\\\\v_f=\dfrac{6.25kg*(8.95m/s)\hat{i}-0.95kg*(10.3m/s(\hat i)+5.4m/s(\hat j))}{6.25kg}\\v_f=7.38m/s(\hat i)-0.82m/s(\hat j)

The angle is given by:

\theta_b=acrtg(\dfrac{v_f(\hat j)}{v_f(\hat i)})\\\\\theta_b=arctg(\frac{-0.82m/s}{7.38m/s})\\\theta_b=-6.34^o

the angle is -6.34 degrees.

4 0
3 years ago
A gymnast whose weight is 5 10 N hangs from the middle of a bar supported by two vertical strands of rope. What is the tension i
german
A 765 N because when the gravitational pull of the weight balances hypothetically in the range of the atmosphere, your weight is not doubles but the force extends it by an uprange of up to 150%
8 0
3 years ago
If the radius of a star increases by a factor of 6 the surface area of the star increases how many times?
Dennis_Churaev [7]

Answer:

When radius of star increase by 6 factor then area of star will increase by a factor of 36.

Explanation:

As we know that

Area of star A given as

A=4\pi R^2

Where R is the radius of star.

Area of star when radius become 6 times

A'=4\pi (6R)^2

A'=4\pi \times 36\times R^2

So

A'=36A

We can say that when radius of star increase by 6 factor then area of star will increase by a factor of 36.

8 0
3 years ago
Gauss’s law of magnetism states that the net magnetic flux through any closed surface is zero.
vesna_86 [32]

Answer:

True

Explanation:

Gauss's law:

It is one of the Maxwell's equations which are the foundation of Electrodynamics. According to this law magnetic field has zero divergence and magnetic monopole can't exist. Inside a closed surface, the magnetic flux inward at the south pole will be exactly equal to the outward magnetic flux at the north pole of the magnetic dipole. Thus, the net magnetic flux will be zero.

5 0
3 years ago
You estimate the radius of the big wheel to be 19 m
posledela
Explanation:

first, find the circumference of the wheel by using the formula 2(pi)(r):

2(pi)(19) = 119.380521

divide by 25 secs

119.380521/25 = 4.77522083

round to the nearest tenth is 4.8, so the speed is 4.8mm/sec
7 0
4 years ago
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