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Vladimir [108]
4 years ago
11

What will happen if you rub two different materials together?

Physics
2 answers:
Dafna1 [17]4 years ago
6 0
It will cause heat from friction
Ulleksa [173]4 years ago
6 0
Sometimes, electrons can also be tranferred
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2. Imagine now that one tried to repeat the same measurement, but between the time that the warm water was measured and the time
Svetllana [295]

Answer:

The final temperature is 64.28°C.

Explanation:

Imagine now that one tried to repeat the same measurement, but between the time that the 400 cm³ of warm water was measured at 88°C and the time it was mixed with the 130 cm³ of cooler water at 15°C, the warmer water had lost 3060 cal to the environment. Calculate the final equilibrium temperature assuming no heat loss after mixing occurs.

Let the temperature of hot water before mixing is t

We need to calculate the temperature

Using formula of energy

Q=mc\Delta t

Q=V\times\rho\times c\times(T_{2}-T_{1})

Put the value into the formula

3060=400\times1\times1\times(88-t)

3060=400\times88-400t

-t=\dfrac{3060-400\times88}{400}

t=80.35^{\circ}

We need to calculate the final temperature

Using formula of temperature

T_{f}=\dfrac{V_{w}T_{w}+V_{c}T_{c}}{V_{w}+V_{c}}

Put the value into the formula

T_{f}=\dfrac{400\times80.3+130\times15}{400+130}

T_{f}=64.28^{\circ}C

Hence, The final temperature is 64.28°C.

5 0
4 years ago
A 3kg chunk of putty moving at a speed of 5 m/a collides with, and sticks to, a 5kg. bowling ball that is initially at REST. The
astraxan [27]
You'd have an easier time using the equation if you understood where the equation comes from.

The law here ... the major principle to remember, the key, the fundamental truth, the big cookie ... is the fact that momentum is conserved.  <em>The total momentum after they join up is the same as the total momentum before they meet.</em>

Momentum of an object is (mass) times (speed).

Now, list all the things you know, before and after the putty meets the ball:

<u>Before:</u><u>  </u>There are two objects.
Mass of putty = 3 kg
Speed of putty =  5m/s
Momentum of putty = 3 x 5 = 15 kg-m/s.

Mass of ball = 5 kg
Speed of ball = zero
Momentum of ball = 5 x 0  =  zero

Total momentum of both things = 15 kg-m/s


<u>After</u>:  There is only one object, because they stuck together.
Mass of (putty+ball) = (3+5) = 8 kg
Speed of (putty+ball) = we don't know; that's what we have to find

Momentum of (putty+ball) = 8 x (speed)

===================================

We know that the momentum after is equal to the momentum before.

8 x (speed) = 15 kg-m/s

Divide each side by  8 :

Speed = 15 / 8  =  <em>1.875 m/s </em>  after they stick together.

4 0
3 years ago
The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is
77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

\% Percent = \frac{V_{nucleus}}{V_{atom}}*100

\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100

\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100

\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100

\% Percent = 7.51*10^{-13}*100

\% Percent = 7.51*10^{-11}\%

Therefore the percent of the atom's volume is occupied by mass is 7.51*10^{-11}\%

3 0
3 years ago
Meghan and Kailey are attempting to push a piano down the hallway, but as they turn up the 8thgrade hallway they are faced with
Alex Ar [27]

Answer: GIVE ME MIN

Explanation:

6 0
3 years ago
Read 2 more answers
Four masses are located in the xy plane. A 4.00 kg mass is located at the origin, and a second 4.00 kg mass is located at the po
Sergio [31]

Answer:

A)    I_total = 16 m, B)   I_total = 8 m,  C)   I_total = 8 m, D)  I_total = 8 m

Explanation:

The moment of inertia is a scalar quantity, therefore the total moment of inertia

          I_total = I₁ + I₂ + I₃ + I₄

the moment of inertia of a point mass with respect to an axis of rotation

         I = m r²

Let's apply this to our case

A) Rotation axis at the origin

     I₁ = m 0 = 0

for the second masses, we find the distance using the Pythagorean theorem

     r = \sqrt{2^2 + 2^2}

     r = 2 √2

     I₂ = m (2 √2) ²

     I₂ = 8 m  

     I₃ = m 2² = 4 m

     I₄ = m 2² = 4 m

we substitute

     I_total = 0 + 8m + 4m + 4m

     I_total = 16 m

B) axis of rotation in the center of the square

let's find the distance to any mass

     r = \sqrt{1^2+ 1^2}

     r = √2

     I₁ = m 2

     I₂ = m 2

     i₃ = m 3

     I₄ = m 4

we substitute

    I_total = 4 (2m)

    I_total = 8 m

C) axis of rotation is the x axis

       I₁ = 0

       I₂ = m 2² = 4 m

       I₃ = m 2² = 4 m

       I₄ = 0

       I_total = 8 m

D) axis of rotation is the y-axis

       I₁ = 0

       I₂ = 4m

       I₃ = 0

       I₄ = 4 m

       I_total = 8 m

5 0
3 years ago
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