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n200080 [17]
2 years ago
7

Help on matching please, struggling on it. Even just 1 or 2 would help

Physics
1 answer:
n200080 [17]2 years ago
6 0

Answer:

7. free fall -- h. 9.8m/s^2

3. Velocity -- x. 60 km/hr west

6. Acceleration -- d. change in velocity/time

8. Centrifugal --  s. towards the centre

13. Work done --w. Force * displacement

5. Uniform circular motion --j. spin cycle in washer

18. Power -- r. kW an hour

7. g -- a. 10N

hope this helps

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Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev
Yanka [14]

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

6 0
2 years ago
Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col
Ratling [72]
In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
p_f=(m+M)v_f=4.8 kg m/s
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find v_f, the velocity of the nail and the block after the collision:
v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}=  0.48 m/s

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
p_i = mv_i = p_f
Rearranging the formula, we can find v_i, the velocity of the nail before the collision:
v_i =  \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s
6 0
3 years ago
Read 2 more answers
Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they
maw [93]
 <span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is: 

Ff = UsN 

where Us is the coefficient of static friction and N is the normal force. 

In order to get the crate moving you must first apply enough force to overcome the static friction: 

Fapplied = Ff 

Since Fapplied = 43 Newtons: 

Fapplied = Ff = 43 = UsN 

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
7 0
3 years ago
Read 2 more answers
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe
stiks02 [169]

Answer:

6.0 × 10^{11} W/m^{2}

Explanation:

From Wien's displacement formula;

Q = e AT^{4}

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = \frac{Q}{A} = eT^{4}

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × (1000)^{4}

                             = 0.6 × 1.0 × 10^{12}

                             = 6.0 × 10^{11} \frac{W}{m^{2} }

Therefore, the emissive intensity coming out of the surface is 6.0 × 10^{11} W/m^{2}.

3 0
2 years ago
The humber bridge in england has the world's longest single span, 1410 m . calculate the change in length of the steel deck of t
Pepsi [2]
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length

Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m
3 0
2 years ago
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