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astra-53 [7]
2 years ago
7

A 4.0 kg block is resting on a rough horizontal table. The coefficient of static friction us is 0.60. The static friction betwee

n the block and the table before any attempted motion is to start is N. (round to the nearest integer)
Physics
1 answer:
professor190 [17]2 years ago
6 0

Answer: 24 N

Explanation:

Given that the

Mass m = 4 kg

Coefficient of friction (μ) = 0.60

To calculate the static friction between the block and the table before any attempted motion, you will use the formula; F = (μ)N

Where

F = static friction

N = normal reaction = mg

g = acceleration due to gravity = 9.8 m/^2

Substitutes m, g and (μ) into the formula

F = 0.6 × 4 × 9.8

F = 23.52 N

Therefore, the static friction between the block and the table before any attempted motion is 24 N approximately.

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The two major problems with most motor vehicles are that they burn fossil fuels and _____________.
Citrus2011 [14]

Answer:

A. Create radioactive waste i believe

Explanation:

8 0
2 years ago
Read 2 more answers
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0
2 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a
Korvikt [17]

Answer:

v^2=v_o^2-2\times a\times (v_o.t)

Explanation:

Given:

Initial velocity of the vehicle, v_o

distance between the car and the tree, x_f

time taken to respond to the situation, t

acceleration of the car after braking, a

Using equation of motion:

v^2=u^2+2a.s ..............(1)

where:

v= final velocity of the car when it hits the tree

u= initial velocity of the  car when the tree falls

a= acceleration after the brakes are applied

s= distance between the tree and the car after the brakes are applied.

s=v_o\times t

Now for this situation the eq. (1) becomes:

v^2=v_o^2-2\times a\times (v_o.t) (negative sign is for the deceleration after the brake is applied to the car.)

5 0
2 years ago
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