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muminat
3 years ago
13

Ice of mass 5 g at 0 °C melts to water at 0 °C.

Physics
1 answer:
amid [387]3 years ago
5 0

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

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a person when asked to speak up,increases her sound level from 30dB to 60dB.The amount of power per unit area increased by? a)30
aivan3 [116]

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

L_2 - L_1 = 10 Log \frac{I_2}{I_1}

so here we need to find the ratio of two intensity

it is given as

Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}

Log\frac{I_2}{I_1} = \frac{60 - 30}{10}

Log\frac{I_2}{I_1} = 3

now we have

\frac{I_2}{I_1} = 10^3

so it is

d) 1000 times

4 0
3 years ago
You travel in a car 30 miles north to ava and home again. what is your distance? __________________________________________ what
Luden [163]
If you travel 30 miles somewhere and then come home again your distance is 60 miles.  Your displacement is 0 because it is the ending position minus the beginning position, which are the same place (home).  In other words, displacement is a vector and distance is a scalar.
4 0
3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti
777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }

a=-0.9408 \frac{m}{s^{2}}

v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m

4 0
3 years ago
What is the total energy equation?
netineya [11]
The total energy equation would be Kinetic energy+Potential energy
6 0
4 years ago
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