A football player collides with another player. The force of the collision is
1000 kg•m/s. The collision lasts 1 second. What is the impulse?​

Answer:

98N

Explanation:

El peso se mide en kg y la fuerza no afecta

Answer

One molecule of water

Answer:

the number of photons absorbed by the solar panel is 1.08 x 10²¹

Explanation:

Given;

frequency of each photon absorbed, f = 5045 x 10¹⁴ Hz

energy to be created by the solar panel, E = 360 kJ = 360,000 J

The energy of each photon absorbed is calculated as;

[tex]E_{photon} = hf\\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E_{photon} = (6.626 \times 10^{-34} )(5045 \times 10^{14})\\\\E_{photon} = 3.343 \times 10^{-16} \ J[/tex]

let the number of photons absorbed = n

[tex]n(E_{photon}) = 360,000 \ J\\\\n = \frac{360,000 \ J}{3.343 \times 10^{-16} \ J} \\\\n = 1.08 \times 10^{21} \ photons[/tex]

Therefore, the number of photons absorbed by the solar panel is 1.08 x 10²¹

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is

43.8 N = k (0.155 m)   ==>   k = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

Explanation:

The magnetic field strength is needed to generate a peak voltage is 0.287 T.

The given parameters;

The maximum voltage in the coil is calculated as follows;

[tex]E_{max} = NAB\omega \\\\[/tex]

where;

The magnetic field strength is needed to generate a peak voltage is calculated as;

[tex]B = \frac{E_{max} }{NA \omega } \\\\B = \frac{E_{max} }{N (\frac{\pi d^2}{4} ) \times 2\pi f}\\\\ B = \frac{330}{120 \times (\frac{\pi \times 0.18^2}{4} ) \times 2\pi \times 60} \\\\B = 0.287 \ T[/tex]

Thus, the magnetic field strength is needed to generate a peak voltage is 0.287 T.

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Answer:

Option C

Explanation:

C. Radiation.....

Answer:

heat travel through vacuum by radiation

Answer:

D) 0.008

Explanation:

I just did it on Edg22 and I got it correct :D (Picture above)

Option D) 0.008T is the correct answer.

Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T

Given the data in the question;

First scenario

Second scenario

Magnetic field is a vector field or region around a magnet or electric charge upon which magnetic force is exerted.

To determine the strength of the magnetic field if the current is changed, we equate the two scenario.

[tex]\frac{B}{I} = \frac{B'}{I'}[/tex]

We substitute our given values into the expression

[tex]\frac{0.004T}{20A} = \frac{B'}{40A}\\ \\ B' * 20A = 0.004T * 40A\\\\B' = \frac{0.004T * 40A}{20A} \\\\B' = \frac{0.004T * 40}{20}\\ \\B' = \frac{0.16T}{20}\\ \\B' = 0.008T[/tex]

Option D) 0.008T is the correct answer.

Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T

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Answer:

45m/s

Explanation:

(here I'm taking gravity due to acceleration as 10 m/s^2 and not 9.8 m/s^2 to match the options)

We know that when an object is dropped from a certain height it has initial velocity 0 m/s

so;

u= 0m/s

s= 100m

a= 10m/s^2

Using 4th law of motion;

v^2 - u^2 = 2× a× s

v^2 - 0^2 = 2 × 10 × 100

v^2 = 2000

v = √2000

v = 44.72 m/ s

Final velocity = 45 m/ s (rounding off to 45)

Please feel free to tell me if you have any confusion.

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

Answer:

(m1+m2)vo=m1v+m2×0

⟹vo=m1+m2m1v

The impulse imparted to second object is equal to change is momentum is -

J=m2vo=m1+m2m2m1v

Answer:

the acceleration of the small bug is 12.83 m/s²

Explanation:

Given;

time of motion, t = 0.5 s

radius of the circular path created by his arm, r  = 1.3 m

if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰

The centripetal acceleration of the ball is calculated as;

[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]

[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]

Therefore, the acceleration of the small bug is 12.83 m/s²

Answer:

The required emf moved across the wire is zero

Explanation:

For a moving charge particle, the magnetic force can be determined by using the formula;

[tex]\varepsilon = Bvlsin \theta[/tex]

since the wire moves in parallel, the angle [tex]\theta[/tex] between magnetic field and velocity = 0°

B = 0.500 T

v = 1.50 m/s

l = 0.200 m

∴

[tex]\varepsilon = (0.500 \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)[/tex]

[tex]\varepsilon = 0.15\times sin (0)[/tex]

[tex]\varepsilon = 0[/tex]

Answer:

[tex]F=1152N[/tex]

Explanation:

From the question we are told that:

Mass/one sec [tex]m=36.0kg/s[/tex]

Incident water stream [tex]v=16.0m/s[/tex]

Exiting water stream [tex]v'=-16.0m/s[/tex]

Generally the equation for Force is mathematically given by

[tex]F=\frac{mv-mu}{t}[/tex]

[tex]F=\frac{36(-16-16)}{1}[/tex]

[tex]F=1152N[/tex]

Answer:

true ,the kinetic energy is associated with energy

Answer:

The weight of the planet is 29083.5 N .

Explanation:

mass of satellite, m = 6463 kg

height of orbit, h = 4.82 x 10^5 m

period, T = 2 h

radius of planet, R = 4.29 x 10^6 m

Let the acceleration due to gravity at the planet is g.

[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]

The weight of the satellite at the surface of the planet is

W = m g = 6463 x 4.5 = 29083.5 N

Answer:

ocean covers 71 percent of the earth

Answer:

the ocean  covers 71 percent of Earth's surface.

Explanation:

Explanation:

Given that,

Three resistors 4ohms 6ohms 8ohms are connected in series and are connected to cell of EMF 60volt.

The equivalent resistance in series combination is given by :

R = R₁ + R₂ +R₃

Put all the values,

R = 4 + 6 + 8

R = 18 ohms

Let I is the current in circuit. So,

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{60}{18}\\\\I=3.33\ A[/tex]

Potential difference in 4 ohms,

[tex]V_1=IR_1\\\\V_1=3.33\times 4\\\\=13.32\ V[/tex]

Potential difference in 6 ohms,

[tex]V_2=IR_2\\\\V_2=3.33\times 6\\\\=19.98\ V[/tex]

Potential difference in 8 ohms,

[tex]V_3=IR_3\\\\V_3=3.33\times 8\\\\=26.64\ V[/tex]

Terminal potential difference in circuit is :

V = IR

Put all the values,

V = 3.33 × 18

V = 59.94 volts

Hence, this is the required solution.

Answer:

[tex]\mu = 3.36\times 10^{-3}\ A-m^2[/tex]

Explanation:

Given that,

The magnitude of magnetic field, B = 0.55 T

The radus of the loop, r = 43 cm = 0.43 m

The current in the loop, I = 5.8 mA = 0.0058 A

We need to find the magnetic moment of the loop. It is given by the relation as follows :

[tex]\mu = AI\\\\\mu=\pi r^2\times I[/tex]

Put all the values,

[tex]\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2[/tex]

So, the magnetic moment of the loop is equal to[tex]3.36\times 10^{-3}\ A-m^2[/tex].

Answer:

The torque is 0.31 Nm.

Explanation:

Electrical energy, E = 8400 J

time, t = 1 min

Angular speed, w = 2900 rpm = 303.53 rad/s

efficiency = 2/3 of input power

The toque is given by  

[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]

Explanation:

The inductive reactance [tex]X_L[/tex] is given by

[tex]X_L = \omega L = 2 \pi fL[/tex]

Solving for f, we get

[tex]f = \dfrac{X_L}{2 \pi L} = \dfrac{12000\:\text{ ohms}}{2\pi (0.8\:H)}[/tex]

[tex]\:\:\:\:\:\:\:= 2387.3\:\text{Hz}[/tex]

Answer:

4 secs

Explanation:

The first step is to calculate the velocity

V= frequency × wavelength

= 500× 0.2

= 100

Therefore the time can be calculated as follows

= distance/velocity

= 400/100

= 4 secs

Solution :

We know, resistance is given by :

[tex]R = \dfrac{\rho l}{A}[/tex]

[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]

Now, we know mass of wire is given by :

[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]

Hence, this is the required solution.

Given:

The resistance will be:

→ [tex]R = \frac{\rho l}{A}[/tex]

or,

→ [tex]A = \frac{\rho l}{R}[/tex]

By substituting the values, we get

      [tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]

      [tex]= 4.816\times 10^{-7} \ m^2[/tex]

hence,

The mass will be:

→ [tex]Mass = Density\times Volume[/tex]

             [tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]

             [tex]= 0.01500 \ kg[/tex]

             [tex]= 15.00 \ g[/tex]

Thus the above answer is right.  

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Answer:

60MW wasted

Explanation:

600-540

=60MW

Answer:

The full form of MBBS is Bachelor of Medicine, Bachelor of Surgery

pls mark me brainliest :))

The full form of MBBS in India is 'Bachelor of Medicine, Bachelor of Surgery'. However, MBBS is an abbreviation of Medicinae Baccalaureus Baccalaureus Chirurgiae, which is the term used for this course in Latin.

Answer:

Sound is produced when an object vibrates, creating a pressure wave. This pressure wave causes particles in the surrounding medium (air, water, or solid) to have vibrational motion. As the particles vibrate, they move nearby particles, transmitting the sound further through the medium.

Answer:

[tex]m=1.47\times 10^{-24}\ Kg[/tex]

Explanation:

Given that,

Charge, [tex]q=1.6\times 10^{-19}\ C[/tex]

Revolution = 7 rev

magnetic field, B = 45 mT

Time, t = 1.29 ms

We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :

[tex]m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg[/tex]

So, the mass of the ion is equal to [tex]1.47\times 10^{-24}\ Kg[/tex].