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ivolga24 [154]
3 years ago
12

What happens to the mass of a metal ball if its heated?

Physics
2 answers:
Lena [83]3 years ago
6 0
Nothing it will turn into a liquid and it will not weigh anything
erik [133]3 years ago
5 0
<span>Mass doesn't change when the temperature of the ball changes.
(Unless, of course, it gets so hot that it melts, and part of it falls off
and rolls under the table.)</span>
You might be interested in
A long point-object, mass = 1.0 kg, moves in a circular path at a radial distance = 0.5 m from the axis of rotation. What is the
Vinvika [58]

Answer:0.25\ kg-m^2

Explanation:

Given

mass of Point object m=1 kg

Distance r=0.5 m

Since mass is moving in circular path therefore every time mass is at distance of r from center .

Also Moment of Inertia tells about  the distribution of mass over the given region with respect to center of mass.

Therefore I=mr^2

I=1\times 0.5^2

I=0.25\ kg-m^2

5 0
3 years ago
Capacitances of 10uF and 20uF are connected in parallel,
jarptica [38.1K]

Answer:

The equivalent capacitance will be 15\mu F  

Explanation:

We have given two capacitance C_=10\mu F\ and\ C_2=20\mu F

They are connected in parallel

So equivalent capacitance C=C_1+C_2=10+20=30\mu F

This equivalent capacitance is now connected in series with 30\mu F

In series combination of capacitors the equivalent capacitance is given by \frac{1}{C}=\frac{1}{30}+\frac{1}{30}

C=\frac{30}{2}=15\mu F

So the equivalent capacitance will be 15\mu F  

5 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
3 years ago
Physics Kinematics question
just olya [345]
An interesting problem, and thanks to the precise heading you put for the question.

We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.

Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m

Assume g = -9.81 m/s^2

initial velocity, v m/s (to be determined)

Solution:

(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation, 
S(T)=800+(vy)T+(1/2)aT^2 ....(1)  
Where S is height measured from ground.

substitute values in (1):  S(20)=800+(0.8v)T+(-9.81)T^2  =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s  for T=20 s

(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s

Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m

(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m

(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T => 
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)

vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s  (magnitude)
in direction theta = atan(43.575,138.1) 
= 17.5 degrees with the vertical, downward and forward. (direction)
4 0
3 years ago
A 6.50-g sample of copper metal at 25.0°C is heated by the addition of 84.0 J of energy. The final temperature of the copper is
Mekhanik [1.2K]

Answer:

Final temperature of the copper is 59 degrees Celsius

Explanation:

It is given that,

Mass of the sample of copper metal, m = 6.5 g

Initial temperature of the metal, T_i=25^{\circ}\ C=298\ K

Heat generated, Q = 84 J

The specific heat capacity of liquid water is 0.38 J/g-K

Let T_f is the final temperature of the copper. It can be calculated using the definition of specific heat of any substance. It is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{Q}{mc}+T_i

T_f=\dfrac{84}{6.5\times 0.38}+298

T_f=332\ K

or

T_f=59^{\circ}C

So, the final temperature of the copper is 59 degrees Celsius. Hence, this is the required solution.

4 0
3 years ago
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