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3 years ago

5

g typical coffee mug holds up to 12 oz (340 grams) of water. How much heat energy is required to warm enough water from 23⁰C to

95⁰C for a mug of hot tea?How many kWh would it take to heat this water with an electric stove?

1 answer:

Kryger [21]3 years ago

8 0

Answer:

Explanation:

Heat required to raise the temperature

= mass x specific heat x rise in temperature

= .34 x 4200 x ( 95 - 23 )

= 102816 J .

1 kWh = 1000 x 60 x 60 J

= 3600000 J

102816 J = 102816 / 3600000

= .02856 kWh .

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A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

RSB [31]

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

5 0

3 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

1 year ago

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The

Goshia [24]

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$ .

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$

4 0

3 years ago

When a hot metal cylinder is dropped into a sample of water, the water molecules

Rashid [163]

Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

4 0

3 years ago

A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction

alukav5142 [94]

The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
Therefore, clearing the normal force we have:
The friction is 565N.
(565 / 0.8) = 706.25N. weight.

6 0

4 years ago