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Evgen [1.6K]
3 years ago
14

an force of 5N is applied to a 10 kg object over a time of 3s. Determine the impulse applied to the object ​

Physics
1 answer:
ddd [48]3 years ago
7 0

Answer:

15Ns

Explanation:

Given parameters:

Mass of the object  = 10kg

Force impact  = 5N

Time of action  = 3s

Unknown:

Impulse applied on the object  = ?

Solution:

Impulse is the force that acts on object within a period of time.

 Mathematically;

             Impulse  = Force x time

Now insert the parameters and solve;

           Impulse  = 5 x 3 = 15Ns

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A person pushes a refrigerator across a horizontal floor. The mass of the refrigerator is 110 kg, the coefficient of static fric
Law Incorporation [45]

Answer:

Explanation:

mass of refrigerator, m = 110 kg

coefficient of static friction, μs = 0.85

coefficient of kinetic friction, μk = 0.59

(a) the minimum force required to just start the motion in refrigerator

F = μs x mg

F = 0.85 x 110 x 9.8

F = 916.3 N

(b) The force required to move the refrigerator with constant speed

F' = μk x mg

F' = 0.59 x 110 x 9.8

F' = 636.02 N

(c) Let a be the acceleration.

Net force = Applied force - friction force

F net = 950 - 636.02

F net = 313.98 N

a = F net / mass

a = 313.98 / 110

a = 2.85 m/s²

4 0
3 years ago
A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
E) the time it takes the car to reach 20.0ms-1​
Galina-37 [17]
4.2 thats the answer
8 0
3 years ago
Find :
Pepsi [2]

Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

a) h =  v₀t + ½gt² = 40(4) +  ½(-10)4² = 80 m

or

v² = u² + 2as

h = (0² - 40²) / 2(-10) = 80 m

8 0
3 years ago
A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
  • u = 48/16
  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
  • M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0
3 years ago
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