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Tomtit [17]
3 years ago
6

What force is required to accelerated a body with a mass of 15 kilograms at a rate of 8 m/s2

Physics
1 answer:
rjkz [21]3 years ago
7 0

Force  =  (mass) · (acceleration)

           = (15 kg) · (8 m/s²)

           =  120 kg-m/s²  =  120 newtons
You might be interested in
Define the types of friction and give FOUR examples of each
Verdich [7]

Answer:

Static Friction - acts on objects when they are resting on a surface

Sliding Friction -  friction that acts on objects when they are sliding over a surface

Rolling Friction - friction that acts on objects when they are rolling over a surface

Fluid Friction - friction that acts on objects that are moving through a fluid

Explanation:

Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book , car parked on a hill.

Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.

Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.

Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon,  sucking water through a straw, submarine moving through water.

4 0
3 years ago
Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C
Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo  Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

6 0
3 years ago
A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i
Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2   Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft     and      d(5) = 35 ft

V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft     and      d(2.1) = 8.61 ft

V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s

4 0
3 years ago
In which of the following conditions would there be the most resistance?
Natasha2012 [34]

Answer:

A) A warm wire

Explanation:

A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.

8 0
3 years ago
A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th
Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e)  T = 2π √m / k

h)   a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

          Fe = m a

          -k x = ma

           a = -k / m x

c) the acceleration is

         a = d²x / dt²

     

      If x = A cos wt

        v = dx / dt = -A w sin (wt +Ф)

        a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

        d²x / dt² = -k / m x

   

We call

       w² = k / m

e) substitute to find w

     -A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

      w² = k / m

Angular velocity and frequency are related

       w = 2π f

       f = 1 / T

       

 We substitute

      T = 2π / w

      T = 2π √m / k

g)    v= - A w sin (wt+Ф)

h) acceleration is

       a = - A w² cos (wt+Ф)

8 0
3 years ago
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