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e-lub [12.9K]
3 years ago
10

How many seconds are there in 3 days 38 hrs? Explain.

Physics
1 answer:
MArishka [77]3 years ago
6 0
259200 seconds in 3 days and 136800 seconds in 38 hours
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Ted lifts a 10N weight at a height of 1.5 m in 1
masha68 [24]

Answer:

D. Ted expanded more power.

Explanation:

Given the following data;

For Ted.

Force = 10N

Height = 1.5m

Time = 1 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/1

Power = 15 Watts.

For Johnny.

Force = 10N

Height = 1.5m

Time = 2 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/2

Power = 7.5 Watts

Therefore, from the calculations we can deduce and conclude that Ted expanded more power.

5 0
3 years ago
A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

8 0
3 years ago
Runner x starts 50 feet in front of runner z and moves with a velocity of 8.0 m/s. At what speed would runner z have to run to c
Mariulka [41]

Answer:

15.625

Explanation:

8 0
3 years ago
4
vfiekz [6]

Answer:

B. fjords

explanation :

Fjords were created by glaciers. In the Earth's last ice age, glaciers covered just about everything. Glaciers move very slowly over time, and can greatly alter the landscape once they have moved through an area. This process is called glaciation.The fjords are one of the glaciers that existed in the past. They are one of the many glacial relief forms that can give us a insight into the size and power of the glaciers.

5 0
3 years ago
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su
CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0
3 years ago
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