A sample of water with a mass of 1 gram freezes at 32°F. At what temperature will a 100-gram sample of water freeze?

Object A is 71 degrees and object B is 75 degrees how will thermal energy flow

Tasya [4]

Given :

Object A is 71 degrees and object B is 75 degrees .

To Find :

How will thermal energy flow.

Solution :

We know, by law of thermodynamics thermal energy will flow from higher temperature to lower temperature.

So, in the given question energy will flow from object B from object A.

Hence, this is the required solution.

3 0

2 years ago

What is the difference between resilience and modulus of resilience

pshichka [43]

<span>Resilience is the amount of energy that can be put into a volume of material and still be stored elastically. ie When the energy goes away, the material regains its undeformed shape. The Mod of R is the amount that can be stored by a unit volume of that material. The Mod of R is heavily related to Youngs Modulus.</span>

6 0

3 years ago

If a vector that is 3cm long represents 30 km/h, what velocity does a 5 cm long vector which is drawn using the same scale repre

FrozenT [24]

Answer:

False knowww yess ofcure

8 0

3 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago

Convection currents in air and water occur because _

Diano4ka-milaya [45]

warm fluids are less dense than cold fluids

7 0

3 years ago