What would happen to the fossil when weathering occurs. Would it survive ? ASAP

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

<span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

</span>

4 0

2 years ago

Read 2 more answers

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Anastaziya [24]

Answer:

As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.

This will mean that the denser objects will always go to the bottom.

This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.

There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.

The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.

6 0

2 years ago

A lamp hangs from the ceiling at a height of 2.6 m. The lamp has a mass of 3.8 kg. The screws holding the lamp break, and it fal

iVinArrow [24]

Answer:

Explanation:

Given height of lamp from the ceiling = 2.6m

mass of the lamp = 3.8kg

acceleration due to gravity = 9.81m/s²

As the body falls to the ground, it falls under the influence of gravity.

Gravitational potential energy = mass*acc due to gravity * height

Gravitational potential energy = 3.8*2.6*9.81

Gravitational potential energy = 96.923 Joules

b) Kinetic energy = 1/2 mv²

m = mass of the body (in kg)

v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion $v^{2} = u^{2}+2gh$

$v^{2} = 0^{2}+2(9.81)(2.6) \\v^{2} = 51.012\\v =\sqrt{51.012}\\ v = 7.14m/s$

Since mass = 3.8kg

$K.E = 1/2 * 3.8 *7.14^{2}\\ K.E = 96.86Joules$

c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s

4 0

2 years ago

A refrigerator is used to cool water from 23°C to 5°C in a continuous manner. The heat rejected in the condenser is 570 kJ/min a

GenaCL600 [577]

Answer: Q=5.46 L/s

COP=2.58

Explanation:

Given that

Cp = 4.18 kJ/(kg.C

density = 1 kg/L

Heat rejected Qr= 570 kJ/min

Power in put W= 2.65 KW

From first law of thermodynamics

U = W+ q

q = Heat absorbed

U = internal energy

W = workdone

U = 570 kJ/min = 9.5 KW

9.5 = 2.65 + q

q = 6.85 KW

COP = q/W

COP = 6.58 / 2.65

COP=2.58

Lets take volume flow rate is Q

So mass flow rate of water m = ρ Q

q = m Cp ΔT

6.85 = 1 x Q x 4.18 ( 23-5)

Q=0.091 L/min

Q=5.46 L/s

7 0

2 years ago

The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie

pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

$\bf \longrightarrow {v}^{2} = {u}^{2} + 2as \\ \\ \rm \longrightarrow {0}^{2} = {15}^{2} + 2 \times a \times 63 \\ \\ \rm \longrightarrow 0 = 225 + 126a \\ \\ \rm \longrightarrow 126a = - 225 \\ \\ \rm \longrightarrow a = - \dfrac{225}{126} \\ \\ \rm \longrightarrow a = - 1.78 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0

2 years ago