A 0.145 kg baseball is thrown with a velocity of 40 m/s. What is the baseball's momentum?

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

tatyana61 [14]

Answer:

$Q_T=63313.5\ J$

Explanation:

Given:

temperature of skin, $T_s=34^{\circ}C$

initial temperature of steam vapour, $T_v=100^{\circ}C$

latent heat of steam, $L=2256\ J.g^{-1}$

mass of steam, $m=25\ g$

specific heat of water, $c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}$

final temperature, $T_f=34^{\circ}C$

<em>Assuming that no heat is lost in the surrounding.</em>

$Q=m.c.\Delta T$

<u>Now the total heat given by the steam to form water at the given conditions:</u>

$Q_T=Q_{Lv}+Q_w$ ..............................(1)

where:

$Q_{Lv}=$ latent heat given out by vapour to form water of 100°C

$Q_w=$ heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

$Q_T=m(L+c.\Delta T)$

$Q_T=25(2256+4.19\times 66)$

$Q_T=63313.5\ J$

is the heat transferred to the skin.

4 0

3 years ago

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$ , and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$ , then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

$V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V$

(b) 1.6 W

In this case, we have:

$R=9.0 k\Omega = 9000 \Omega$ is the resistance of the resistor

$V=120 V$ is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

$P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W$

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

$R_1 = 100 \Omega\\R_2 = 150 \Omega$

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

$V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V$

While the greatest potential difference allowed in the second resistor is

$V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V$

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

$P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W$

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0

3 years ago

A screw having 50% efficiency is driven by a rod and 25 cm. The pitch of the screw is 1/10cm Calculate velocity ratio and mechan

neonofarm [45]

(a) The velocity ratio of the screw is 1570.8.

(b) The mechanical advantage of the screw is 785.39.

<h3>Velocity ratio of the screw</h3>

The velocity ratio of the screw is calculated as follows;

V.R = 2πr/P

where;

P is the pitch = 1/10 cm = 0.1 cm = 0.001 m
r is radius = 25 cm = 0.25 m

V.R = (2π x 0.25)/(0.001)

V.R = 1570.8

<h3>Mechanical advantage of the screw</h3>

E = MA/VR x 100%

0.5 = MA/1570.8

MA = 785.39

Learn more about mechanical advantage here: brainly.com/question/18345299

#SPJ1

4 0

2 years ago

HURRYYYY pls help due today n i be giving brainliest!!!<br><br><br> n no mfkn links plsss

nordsb [41]

Answer:

1 Frequency

2 Wavelength

3 Amplitude

4 Crest

Hope it helps pls mark brainliest

5 0

3 years ago

A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A

KIM [24]

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

1.6 x = 1.2

x = 0.75

So the bulb should located no more than 0.75 m from the mirror.

5 0

3 years ago