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3 years ago

5

A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma

gnitude of the vertical component of its initial velocity?

2 answers:

tangare [24]3 years ago

5 0

Answer:

8.7

Explanation:

(10)sin60

Flura [38]3 years ago

3 0

Answer:

5 sq. root 3

Explanation:

theta= 60°

=> u sin theta = 10 × sin 60

= 10× sq. root 3/2

= 5 sq. root 3

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A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

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

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

I hope it helps you!

8 0

3 years ago

nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

$a = \frac{vf - vi}{t}$

=

$\frac{15 - 25}{5}$

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0

3 years ago

The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?

Sonja [21]

Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.

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3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

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3 years ago