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2 years ago

True or false. The planet's speed never changes. That is why we have a stable orbit.

Physics

1 answer:

Svetradugi [14.3K]2 years ago

8 0

This answer is true the earth always stays at one speed

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It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

just olya [345]

So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c

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3 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

Equation for pressure at a depth H inside a fluid PLSS URGENTT

Blababa [14]

Answer:

We begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

4 0

3 years ago

1. A toy car with mass m1 travels to the right on a frictionless track with a speed of 3 m/s. A second toy car with mass m2 trav

Alika [10]

Answer:

v = 3(m1 - 2m2)/(m1 + m2)

Explanation:

Parameters given:

Velocity of first toy car with mass m1, u1 = 3 m/s (taking the right direction as the positive axis)

Velocity of second toy car with mass m2, u2 = -6 m/s (taking the left direction as the negative x axis)

Using conservation of momentum principle:

Total initial momentum = Total final momentum

m1*u1 + m2*u2 = m1*v1 + m2*v2

Since they stick together after collision, they have the same final velocity.

m1*3 + (m2 * -6) = m1*v + m2*v

3m1 - 6m2 = (m1 + m2)v

v = (3m1 - 6m2) / (m1 + m2)

v = 3(m1 - 2m2) / (m1 + m2)

8 0

3 years ago

Read 2 more answers

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this

Afina-wow [57]

Answer: $27^{\circ}$

Explanation:

Given

Initial velocity of both snowball is 29.3 m/s

first snowball launch angle $=63^{\circ}$

Considering motion of snowball to be projectile

range is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{29.3^2\sin 126}{9.8}$

R=70.87 m-----1

If second snowball is thrown at an angle of \phi

$R=\frac{u^2\sin 2\phi }{g}$

$R=\frac{29.3^2\sin 2\phi }{9.8}$ ------2

$70.87=87.601\sin 2\phi$

$0.809=\sin \phi$

$2\phi can be 53.99^{\circ}$

or $180-2\phi =53.99^{\circ}$

Thus $\phi =26.995 \approx 27^{\circ}$

4 0

3 years ago