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3 years ago

6

When you walk at an average speed (constant speed, no acceleration) of 20.7 m/s in 75.8 sec you will cover a distance of______?

1 answer:

insens350 [35]3 years ago

7 0

Answer:

You will cover a distance of 1569.06 metres. Or you could round down to 1,569m.

Explanation:

20.7*75.8=1563.06

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A disk rotates freely on a vertical axis with an angular velocity of 50 rpm . An identical disk rotates above it in the same dir

True [87]

Answer:

Final angular velocity is 35rpm

Explanation:

Angular velocity is given by the equation:

I1w1i + I2w2i = I1w1f -I2w2f

But the two disks are identical, so Ii =I2

wf can be calculated using

wf = w1i - w2i/2

Given: w1i =50rpm w2i= 30rpm

wf= (50 + 20) / 2

wf= 70/2 = 35rpm

5 0

2 years ago

What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t

Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0

2 years ago

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea

steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

$m_1 = m_2$

$\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}$

Pressure is same because it's not changing due to process

$\dfrac{V}{500} = \dfrac{2 V}{T_2}$

$T_2 = 1000\ K$

$\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}$

$\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})$

$m = \dfrac{P_1V_1}{RT_1}$

$m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}$

m = 1.39 x 10⁻³ Kg

$\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)$

$\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K$

5 0

2 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

11 months ago