Velocity hdjsksskskskdldlddld

Greg is boiling water on the stove for some noodles. what object can he safely hold while draining the water from the noodles af

zmey [24]

In order for Greg to safely drain the water out of the noodles, he should use potholders or any thing that is does not conduct heat or transfer heat. Some pots are also equipped with handles that are made of plastics for safely transferring of its content to another container.

5 0

3 years ago

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

8 0

3 years ago

Read 2 more answers

-What do you think happened to make the Moon look the way it does?

goldenfox [79]

Answer: The physics of evolution had made the moon like it is today....Please watch this video from you tube about the evolution of the moon.

Explanation:

https://youtu.be/UIKmSQqp8wY

5 0

3 years ago

Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

3 0

3 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago

Velocity hdjsksskskskdldlddld​

Velocity hdjsksskskskdldlddld