Answer:
a) W = 2746.8[J]
b) W = 2992.05 [J]
Explanation:
Work is defined as the product of force by distance. We must bear in mind that the force that performs the work is the one that acts in the same direction of displacement.
For this case, we must calculate the weight of the woman, the weight is defined as the product of mass by gravity.
where:
w = weight [N] (units of Newtons]
m = mass = 56 [kg]
g = gravity acceleration = 9.81 [m/s²]
![w=56*9.81\\w=549.36[N]](https://tex.z-dn.net/?f=w%3D56%2A9.81%5C%5Cw%3D549.36%5BN%5D)
a)
where:
F = weight = 549.36[N]
d = distance = 5 [m]
![W = 549.36*5\\W = 2746.8[J]](https://tex.z-dn.net/?f=W%20%3D%20549.36%2A5%5C%5CW%20%3D%202746.8%5BJ%5D)
b)
The new mass will be the combination of the mass of the woman plus that of the load.
![m_{new} = 56+5\\m_{new}=61[kg]](https://tex.z-dn.net/?f=m_%7Bnew%7D%20%3D%2056%2B5%5C%5Cm_%7Bnew%7D%3D61%5Bkg%5D)
![w_{new}=61*9.81\\w_{new}=598.41[N]](https://tex.z-dn.net/?f=w_%7Bnew%7D%3D61%2A9.81%5C%5Cw_%7Bnew%7D%3D598.41%5BN%5D)
The new work done.
![W =598.41*5\\W=2992.05[J]](https://tex.z-dn.net/?f=W%20%3D598.41%2A5%5C%5CW%3D2992.05%5BJ%5D)
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
Answer:
Explanation:
we know that the time rate of change of velocity is called acceleration mathmetically;
a=Δv/Δt
Δv=a*Δt
a=10 m/s²
t=10 sec
Δv=10*10=100 m/s
Solubility is the ability to dissolve in liquids like water or organic solvents.
Malleability is the ability to bend and be hammered without breaking.
Ductility is when a material can be stretched into wires.
It means that 19.3g of gold is packed into 1cm^3.
