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qaws [65]
3 years ago
15

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

ocity of 4.8 m/s, grabs and holds onto him so that they move off together with a velocity of 2.5 m/s. If the mass of the tackler is 100 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.
Physics
1 answer:
zmey [24]3 years ago
4 0

Answer:

Mass of receiver is 92 kg        

Explanation:

We have given mass of tackler m_1=100kg

Let the mass of receiver is m_2kg

When tackler moving alone velocity is v_i=4.8 m/sec

And when tackler and receiver is together velocity is v_f = 2.5 m/sec

So from conservation of momentum

m_1v_i=(m_1+m_2)v_f

100\times 4.8=(100+m_2)\times 2.5

2.5m_2+250=480

2.5m_2=230

m_2=92kg

So mass of receiver is 92 kg

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When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

EMF = v.(B x L)

given that

B = 25\mu Tj - 50\mu Tk

v = 2 m/s j

L = 0.50 m (-i)

now by using the above formula we will have

EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)

EMF = 2(j) .(12.5\mu k + 25\mu j)

EMF = 50 \mu Volts

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3 years ago
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A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /
Step2247 [10]
Va ser 0.0900 yo creo preo que esta respuesta te ayude
5 0
3 years ago
A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration
galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

5 0
3 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

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2 years ago
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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

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a = 6 + 0.02s

so,

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Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
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