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Lera25 [3.4K]
3 years ago
8

Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct

ed south. The net force on the object isA. 8 N south.B. 12 N south.C. 20 N north.D. 8 N north.
Physics
1 answer:
xeze [42]3 years ago
7 0

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

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A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
DiKsa [7]

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

3 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
We attach two blocks of masses m1 = 7 kg and m2 = 7 kg to either end of a spring of spring constant k = 1 N/m and set them into
SVETLANKA909090 [29]

Answer:

The angular frequency \omega of the oscillation is 0.58s^{-1}

Explanation:

For this particular situation, the angular frequency of the system is given by

\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}

6 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
zhuklara [117]

Complete Question

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.681 m apart. At what distance, in kilometres, are you marginally able to discern that there are two headlights rather than a single light source?Take the wavelength of the light to be 549 nm and your pupil diameter as 4.63 mm.  

Answer:

The distance is  z  = 4707.6 \ m

Explanation:

From the question we are told that

    The is distance between the headlight is d = 0.681 \ m

   The wavelength is  \lambda = 549 \ nm = 549 *10^{-9} \ m

    The  pupil diameter is  D  = 4.63 \ mm = 0.00463 \ m

Generally, we can mathematically evaluate the resolution of the eye as

            \theta  = \frac{1.22 *  \lambda }{D}

    substituting values

              \theta  = \frac{1.22 *  549 *10^{-9} }{0.00463}

              \theta  =  (1.45 *10^{-4} )^o

Now according to SOHCAHTOA rule  

         sin \theta  =  \frac{ d}{z}

Where z is  the distance at which the eye can discern the two head light

  given that the angle is very small sin \theta =  \theta

=>    \theta  =  \frac{ d}{z}

substituting values

     1.45*10^{-4}  =  \frac{ 0.681}{z}

=>   z  = \frac{0.681}{1.45 *10^{-4}}

=>    z  = 4707.6 \ m

   

6 0
4 years ago
A heavier mass m1 and a lighter mass m2 are 17.0 cm apart and experience a gravitational force of attraction that is 9.70 10-9 N
puteri [66]

Answer:

The value of m1 is 4.2 kg and value of m2 is 1 kg.

Explanation:

Note:- The value of force is assumed to be 9.70*10^-9 N.

The gravitational force F is given by the formula,

F=Gm1m2 / r^2

where F is the force, G is the universal gravitational constant, m1 and m2 are masses and r is the distance between the masses.

Then the product of the masses is,

m1m2= Fr^2/G

Given F=9.70*10^-9 N, r is 17.0 cm which is equal to 0.17 m.

Therefore product m1m2= 9.70*10^-9 N *(0.17 m)^2 / 6.67*10^(-11) N m^2 kg^-2

m1m2= 4.20 kg^2

m1=4.20 kg^2/m2

The sum of masses is m1+m2=5.20 kg

 4.20/m2+m2=5.20

(m2)^2-5.20m2-4.20=0

After solving the above quadratic equation, mass m2 has values 4.2 kg and 1 kg.

By using m2=1 kg, m1 has greater mass 4.2 kg.

Check out similar question.

brainly.com/question/13014873

#SPJ10

4 0
2 years ago
Read 2 more answers
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