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2 years ago

10

Where are you MOST likely to find platinum? Check all that apply.

1 answer:

saul85 [17]2 years ago

6 0

Answer:

a b and d

Explanation:

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o

Colt1911 [192]

Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

$\beta_c = 51*10^{-6}/\°C$

$\beta_l = 400*10^{-6}/\°C$

$V_0 = 16L$

$\Delta T = (95\°C-10\°C)$

Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

6 0

3 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

3 years ago

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

Angelina_Jolie [31]

Answer:

E1 = 2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

= 2996.667N/C

E2 = kQ2/r^2

= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

= 11237.5N/C

The direction are towards the point a

6 0

3 years ago

The landing of a spacecraft is cushioned with the help of airbags. During its landing on Mars, the velocity of downward fall is

statuscvo [17]

You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same

5 0

3 years ago

Read 2 more answers

2. the energy of the pendulum as it swings up

Nonamiya [84]

kinetic energy or potential energy

sorry idek i learned this like 2 years ago

4 0

3 years ago