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maria [59]
3 years ago
14

A 35kg cannonball sits at rest on a flat level surface. What is the normal force exerted on the cannon ball. Use -9.8 m/s2 for a

cceleration due to gravity.
A. -343N

B. 44.8N

C. 25.2N

D. 343N
Physics
1 answer:
Tom [10]3 years ago
8 0

Answer:

a) N = 343 [N]

Explanation:

We must remember Newton's third law, which tells us that the force acting on a body is equal to the normal reaction force of equal magnitude but acting in the opposite direction.

N=m*g

where:

m = mass = 35 [kg]

g = gravity acceleration = 9.81 [m/s²]

N = 35*9.8\\N= 343 [N]

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A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ
creativ13 [48]

Answer:

Part a)

E = \frac{8.60}{2.62} = 3.28 J

Part b)

E = 2.62(8.60) = 22.5 J

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

U = \frac{Q^2}{2C}

As we know that initial energy is given as

8.60 = \frac{Q^2}{2C}

now we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0A}{d}

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

C' = \frac{Cd}{d'}

C' = \frac{C(3.80)}{1.45}

C' = 2.62 C

Now the new energy of the capacitor is given as

E = \frac{Q^2}{2(2.62C)}

E = \frac{8.60}{2.62} = 3.28 J

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

U = \frac{1}{2}CV^2

8.60 = \frac{1}{2}CV^2

now when capacitance is changed to new value then new energy is given as

E = \frac{1}{2}C'V^2

E = \frac{1}{2}(2.62C)V^2

E = 2.62(8.60) = 22.5 J

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3 years ago
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They are malleable and lustrous, and can conduct both electricity and thermal heat
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3 years ago
A combination of two or more simple machines is a (n)
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A compound Machine is 2 machines that work together in order to make a task easier.
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The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b
Romashka-Z-Leto [24]

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

4 0
3 years ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft&gt;s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
3 years ago
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