What is released when rocks shift or move along a tectonic plate boundary?

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

2 years ago

An example of a high explosive is what?

nordsb [41]

The answer is Dynamite.
Explosive, any substance or device that can be made to produce a volume of rapidly expanding gas in an extremely brief period. Chemical explosives are of two types; detonating, or high explosives and deflagrating, or low, explosives. Detonating explosives, such as TNT and dynamite, are characterized by extremely rapid decomposition and development of high pressure, whereas deflagrating explosives, such as black and smokeless powders, involve merely fast burning and produce relatively low pressures.

5 0

2 years ago

Read 2 more answers

A 90kg person jumps from a 30m tower into a tub of water with a volume of 5m3 initially at 20°C. Assuming that all of the work d

Kryger [21]

Answer:

T₂ = 20.06 ° C

Explanation:

Given

P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C

Using the formula to determine the final temperature of the water

T₂ = T₁ * P * h / Eₐ * c

The work done of the person to the water

Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²

Eₐ = 49000 N

T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]

T₂ = 20.06 ° C

4 0

3 years ago

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

3 years ago

Object A is placed on top of object B. Object A is the same temperature as object B. How will heat flow between object A and obj

mash [69]

"No heat will flow between object A and object B" is the one among the following choices given in the question that describes how heat will <span>flow between object A and object B. The correct option among all the options that are given in the question is the second option or option "B". I hope it helps you.</span>

6 0

2 years ago

Read 2 more answers