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Otrada [13]
3 years ago
9

 Which of the following levels of classification is the least specific?

Physics
1 answer:
Troyanec [42]3 years ago
4 0
Species is the least specific
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A 5kg book rests on a table. How much force is it exerting on the table?​
valentinak56 [21]

Answer:

5 Kg of Force

Explanation:

3 0
3 years ago
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A motor develops 80 kW of power as it lifts a loaded elevator 20.5 m in 45 s. How much force does the motor exert? *
yuradex [85]

Answer:

F=1.3⋅105N

Explanation:

It's common knowledge.

7 0
3 years ago
What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds
Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N

The net power needed to change the speed of the vehicle is calculated as;

P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W

3 0
3 years ago
How much force to move 18 kg at a rate of 3m/s​
Stolb23 [73]

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

4 0
3 years ago
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees
Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt)  ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1  

70 = Tamb + (100 - Tamb)*e^(-10k)   ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k)   ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)  

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k)   .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0
3 years ago
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