Answer:
original mass of the block of ice is 38.34 gram
Explanation:
Given data
cup mass = 150 g
ice temperature = 0°C
water mass = 210 g
water temperature = 12°C
ice melt = 2 gram
to find out
solution
we know here
specific heat of aluminum is c = 0.900 joule/gram °C
Specific heat of water C = 4.186 joule/gram °C
so here temperature difference is dt = 12- 0 = 12°C
so here heat lost by water and cup are given by
heat lost = cup mass × c × dt + water mass × C × dt
heat lost = 150 × 0.900 × 12 + 210 × 4.186 × 12
heat lost = 12168.72 J
so
mass of ice melt here = heat lost / latent heat of fusion
here we know latent heat of fusion = 334.88 joule/gram
so
mass of ice melt = 12168.72 / 334.88
mass of ice melt is 36.337554 gram
so mass of ice is here = mass of ice melt + ice melt
mass of ice = 36.337554 + 2
mass of ice = 38.337554 gram
so original mass of the block of ice is 38.34 gram
Answer:
b.) Length
Explanation:
The length of the string can be changed by removing it from the slotted bracket and placing it back in. You can change the mass by varying the number of washers on the mass hanger. The amplitude can be changed by varying the starting angle of the pendulum (low, medium, and high angle). sorry if wrong
Answer:
Answers of the Both parts are in the following attachment
Explanation:
Answer:
Newton's third law of motion.
Explanation:
We are told the force needed to throw the full soda can was more than that needed to throw the empty can.
Now, the weight of the full soda can will be more than that of the empty can. Therefore, the full can will demand more force than that of the empty can due to Newton's third law of motion which states that to every action, there is an equal and opposite reaction.
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)
