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lapo4ka [179]
3 years ago
8

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

ure of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, T_i=4^{\circ} C

Final temperature of water, T_f=25^{\circ} C

The specific heat capacity of liquid water is, c=4.184\ J/g\ ^oC

Heat transferred is given by :

Q=mc(T_f-T_i)

Q=710\times 4.184\times (25-4)

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

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The stoplights on a street are designed to keep traffic moving at 26 mi/h. the average length of a street block between traffic ligh
bonufazy [111]

We use the formula,

v = \frac{d}{t}.

Here, v is velocity and its value given 26 mi/h ( in m/s,  \frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

t = \frac{80 m}{11.62m/s } = 6.88 \ s

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s


3 0
3 years ago
If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

5 0
3 years ago
Which formula describes Newton’s second law of motion?
Allushta [10]
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

6 0
3 years ago
Read 2 more answers
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0
3 years ago
E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha
anzhelika [568]

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

6 0
3 years ago
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