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3 years ago

if a metal object was cut on half how much heat will be required to raise both half to a specific temperature

1 answer:

3 0

suppose the mass of a metal object be m and its specific heat capacity be s, and, H joules of heat is required to raise its temperature by t degrees Celsius

Than H is given by relation, $H = mst$

Now if this object is cut in two half's, than mass of half part will also be half i.e. M/2

So heat required to heat the half part will be $= (\frac{m}{2}) st =\frac{mst}{2} = H/2$

Hence, the heat required to raise the half object to a specific temperature will also become half.

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A basketball is thrown straight up in the air. At its peak, it is 0.02 km high and has a velocity of 0 m/s. What is its final ve

Aleks04 [339]

Answer:

20 m/s

Explanation:

Given:

Δy = 0.02 km = 20 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (20 m)

v = 19.8 m/s

Rounded to one significant figure, the final velocity is 20 m/s towards the ground.

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2 years ago

HELP ASAP HELP Two tennis balls of the same mass are served at different speeds: 30 m/s and 60 m/s. Which serve has more kinetic

inna [77]

Answer:

The second ball has four times as much kinetic energy as the first ball.

Explanation:

Kinetic Energy

Is the type of energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

Two tennis balls have the same mass m and are served at speeds v1=30 m/s and v2=60 m/s.

The kinetic energy of the first ball is:

$\displaystyle K_1=\frac{1}{2}m\cdot 30^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 900$

$K_1=450m$

The kinetic energy of the second ball is:

$\displaystyle K_2=\frac{1}{2}m\cdot 60^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3600$

$K_2=1800m$

Being m the same for both balls, the second ball has more kinetic energy than the first ball.

To find out how much, we find the ratio:

$\displaystyle \frac{K_2}{K_1}=\frac{1800m}{450m}$

Simplifying:

$\displaystyle \frac{K_2}{K_1}=4$

The second ball has four times as much kinetic energy as the first ball.

5 0

2 years ago

Read 2 more answers

Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

2 years ago

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

2 years ago

What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

Pressure on the 13 th floor, P₁ = 35 PSI
Distance between each floor, d = 10 ft

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0

1 year ago