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Yuri [45]
3 years ago
7

This was due yesterday...

Physics
2 answers:
likoan [24]3 years ago
6 0

Answer:

the answer is c

Explanation:

water is abiotic

Aleonysh [2.5K]3 years ago
5 0

Answer:

the answer is C

Explanation:

Its the only reaction with abiotic and biotic.

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A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, w
sp2606 [1]

Answer:

8.9 g/cm^3

Explanation:

density = mass/volume

volume = length * width * height

volume = (8.4 cm)(5.5 cm)(4.6 cm)

volume = 212.52 cm^3

mass = 1896 g

density = (1896 g)/(212.52 cm^3)

density = 8.9 g/cm^3

3 0
3 years ago
What is kinetic energy?
Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

3 0
3 years ago
Read 2 more answers
Period T= time of one revolution measured in seconds
ser-zykov [4K]
60 RPM equals one hertz (i.e., one revolution per second, or a period of one second). The SI unit for period is the second.
3 0
3 years ago
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

                      = (14) (9.81) (5.5sin36.9°)

                       = 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

                      = -453.5J

8 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
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