Answer:
1398.12 N
Explanation:
We define the x-axis in the direction parallel to the movement of the truck on and the y-axis in the direction perpendicular to it.
x-components of the ropes forces
T₁x = 615N*cos31°=527.1579 N :Tension in direction x of the rope of the car 1
T₂x= 961 N*cos25°=870.96 N :Tension in direction x of the rope of the car 2
Net forward force exerted on the truck in the direction it is headed (Fnx)
Fnx = T₁x + T₂x
Fnx = 527.1579 N + 870.96 N
Fnx = 1398.12 N
Answer:
Explanation:
Given that,
The work function for silver is 4.73 eV.
We need to find the value of the work function from electron volts to joules.
We know that,
For 4.73 eV,
So, the work function for silver is 
.
Answer:
C.
Explanation:
A person pushing a couch will face resistive force of friction . When resistive force is greater then his force of effort , couch will not move. This force is static friction because the couch is stationary. When the force of effort is increased , magnitude of static friction also increases keeping the couch stationary. The ability of the static friction to increase its magnitude is limited by a maximum value beyond which the couch starts moving. The static friction is then converted into kinetic friction.
In rest of the three cases object is already moving so kinetic friction is in action.
<u>Answer:</u> The specific heat of ice is 2.11 J/g°C
<u>Explanation:</u>
When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.
The equation used to calculate heat released or absorbed follows:
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of ice = 12.5 g
= mass of water = 85.0 g
= final temperature = 22.24°C
= initial temperature of ice = -15.00°C
= initial temperature of water = 25.00°C
= specific heat of ice = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]](https://tex.z-dn.net/?f=12.5%5Ctimes%20c_1%5Ctimes%20%2822.24-%28-15%29%29%3D-%5B85.0%5Ctimes%204.186%5Ctimes%20%2822.24-25%29%5D)
Hence, the specific heat of ice is 2.11 J/g°C
Gravity affet everything and it touches nothing.
Hope this helps!
