m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m (T - ) + m = M ( - T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg
The answer is 2.3. I just answered this question and got it right.
The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.
Answer:
Explanation:
Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.
Since the radius of the track is given as 200 m, then the circumference of the circular track will be
Circumference = 2πr = 2 × 3.14 × 200
So the circumference of the circular track = 1256 m.
So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.
As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.
So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.
Then Displacement = Final-Initial = 256-0= 256 m.
So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.
Tons of stuff!
Footpath erosion, increased usage of travel vehicles, construction of hotels and other attractions, ect ect.
Answer:
Explanation:
Given
initial speed of Launch
=53.7 m/s
Range of Projectile =Maximum height of Projectile
Range is given by R
Maximum height is given by 
![\frac{u^2\sin \theta }{g}\cdot \left [ 2\cos \theta -\frac{1}{2}\right ]=0](https://tex.z-dn.net/?f=%5Cfrac%7Bu%5E2%5Csin%20%5Ctheta%20%7D%7Bg%7D%5Ccdot%20%5Cleft%20%5B%202%5Ccos%20%5Ctheta%20-%5Cfrac%7B1%7D%7B2%7D%5Cright%20%5D%3D0)
as u cannot be zero therefore
