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Elina [12.6K]
3 years ago
14

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

the car
Physics
2 answers:
9966 [12]3 years ago
4 0
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
olga_2 [115]3 years ago
3 0

Answer:

The magnitude is 2000N

Explanation:

To solve this question we are going to use the second Newton law.

The second Newton law states that the acceleration of an object is directly proportional to the net force which acts on the object and inverse proportional to the mass of the object ⇒

a=\frac{F}{m}

Which can also be written as :

F=m.a (I)

Where F is the force

Where m is the mass of the object and where ''a'' is the acceleration.

In the problem, we know that

m=1000kg

a=2\frac{m}{s^{2}}

If we replace in the equation (I) :

F=(1000kg).(2\frac{m}{s^{2}})=2000(kg.\frac{m}{s^{2}})

The unit kg.\frac{m}{s^{2}} is defined as N which is called Newton. N is an unit of force.

Finally, the net force exerted on the car is 2000N

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High-speed stroboscopic photographs show that the head of a 180 g golf club is traveling at 47 m/s just before it strikes a 46 g
inn [45]

Answer:

47 m/s

Explanation:

golf club mass, mc = 180 g

golf ball mass, mb = 46 g

initial golf club speed, vc1 = 47 m/s

final golf club speed, vc2 = 35 m/s

initial golf ball speed, vb1 = 0 m/s

final golf ball speed, vb2 = ? m/s

The total momentum is conserved, then:

mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2

Replacing with data and solving (dimension are omitted):

180*47 + 46*0 = 180*35 + 46*vb2

vb2 = (180*47 - 180*35)/46

vb2 = 47 m/s

7 0
3 years ago
Read 2 more answers
A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
r-ruslan [8.4K]

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

\angle ABC =(180-10) \textdegree=170\textdegree

#Distance traveled in 1.5hrs is;

c=690x1.5\\=1035mi

#Distance traveled in next two hrs:

a=690\times 2\\=1380mi

#Now using the Cosine Rule:

b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi

Hence, the pilot is 2406 miles from her starting position.

4 0
3 years ago
14. Saeed is pulling a 6 kg heavy rock with an upward force of 40 N but does not succeed to lift it up. What is the magnitude of
NemiM [27]

Answer:

58.8 N

Explanation:

The normal force is calculated as equal to the perpendicular component of the gravitational force.

Thus; N = mg

We are given m = 6 kg

Thus;

N = 6 × 9.8

N = 58.8 N

Thus, magnitude of normal force on the rock = 58.8 N

6 0
2 years ago
Which of the following is not an essential feature of scientific explanations?
sineoko [7]

Answer:

D I think I might be wrong its been a while scense I did something like that

4 0
3 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
Read 2 more answers
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