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2 years ago

How can you make a solution saturated

Physics

1 answer:

Step2247 [10]2 years ago

5 0

Answer: The answer is B. Add more solute (took test)

Explanation:

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Which direction will thermal energy flow if you pick up a snowball with your bare hand? Thermal energy will flow from the snowba

Lapatulllka [165]

Answer:

b. Thermal energy will flow from your hand to the snowball.

Explanation:

4 0

2 years ago

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Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.

BigorU [14]

Answer:

Explanation:

Since energy is conserved:

+mgh

⇒u

+2gh

⇒(3)

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

7 0

2 years ago

A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?

uysha [10]

Answer:

163.35

__________________________________________________________

<u>We are given:</u>

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

<u>Kinetic Energy of the object:</u>

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)² [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

7 0

2 years ago

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

2 years ago

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$

$265v_r + 72.8*5.21 = 0$

$v_b = -72.8*5.21/265 = -1.43 m/s$

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0

2 years ago