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guapka [62]
3 years ago
5

PLEASE ANSWER FAST!!

Physics
2 answers:
nataly862011 [7]3 years ago
7 0
I think that it’s false I might be wrong but I want the points
Nostrana [21]3 years ago
6 0
The answer is true, because Newton's first law of motion states that "Any object continues in its state of rest or motion unless it is compelled by a net external force."
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A car accelerates from rest, and travels 400 m in 3.5 seconds. If
zheka24 [161]

Answer:

A car accelerates from rest, and travels 400 m in 3.5 seconds. If

the net force on the car is 12,000 N what is the mass of the car? bzgs dvd d dv dvdvd dhd dbvd

Explanation:

shd dhd bdvd dhdbduhdbdhdbbdceudd f

3 0
3 years ago
Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

Average speed of Prav,

v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0
3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

6 0
3 years ago
Read 2 more answers
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the
ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

5 0
3 years ago
Read 2 more answers
A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum
AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, \omega=33\frac{1}{3} rev/min , radius of the record r=0.1m and the distance between any two successive bumps on the groove as d=1.75mm.

The linear speed of the record in meters per second is:

v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\

#From v above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

Hits/second=v/d    \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199

Hence the bumps hit the stylus at around 199hit/s

8 0
3 years ago
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