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2 years ago

13

Spring tides occur when solar and lunar tides are in sync with each other. On the diagram, identify the two moon phases that cor

respond to a spring tide.

1 answer:

loris [4]2 years ago

5 0

Answer

Sun and Moon's gravity

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

4 years ago

Aliens? do you believe?<br><br>To get the points go into detail :)

zzz [600]

Answer:

Yes

Explanation:

There are so many planets out there that there must be habitable planets if not in our galaxy but the Universe.

Although the chances of advanced life are slim, small primitive life like microbes or sea life may still exist.

7 0

4 years ago

Read 2 more answers

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

Klio2033 [76]

Answer:

The radius of a chamber is 2.36 meters.

Explanation:

Given that,

The outer wall moves at a speed of 2.72 m/s.

Mass of the person, m = 75.1 kg

The person feels a force of 235 N force pressing against his back. The force acting on the person is centripetal force. It is given by the below formula :

$F=\dfrac{mv^2}{r}$

r is the radius of a chamber

$r=\dfrac{mv^2}{F}$

$r=\dfrac{75.1\times (2.72)^2}{235}$

r = 2.36 meters

So, the radius of a chamber is 2.36 meters. Hence, this is the required solution.

7 0

3 years ago

The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about 0.35 AU.

Alina [70]

Answer:

52,360,000km

Explanation:

To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

$0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km$

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

5 0

4 years ago

Read 2 more answers

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago