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Juli2301 [7.4K]
2 years ago
13

The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distanc

e d apart on the same axle. From the angular displacement Δθ of the two bullet holes in the disks and the rotational speed of the disks, we can determine the speed v of the bullet. Find the bullet speed for the following data: d = 70 cm, ω = 850 rev/min, and Δθ = 31.0°.........The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks?
Physics
1 answer:
RUDIKE [14]2 years ago
7 0

Answer:

uhhhh

Explanation:

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An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d
Damm [24]

9 Newtons on the right, because:

Right and left are opposite directions, however, in the same direction.

We know that, opposite directions, the signal is of subtraction. So, you need to make the difference between 22 and 13, you will see that it is 9. Therefore, the Resulting Force will be 9.

6 0
3 years ago
Read 2 more answers
A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and rea
Ugo [173]

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement  a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t  Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s  , t= 30s

θ = (58 rad/s)*(30)

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt  = 290 rad+ 1740 rad + 1201.42 rad

θt  = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

7 0
2 years ago
A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car
anygoal [31]

Answer:h=19.4 m

Explanation:

Given

mass of automobile m=750\ kg

Initial height of automobile h_o=5\ m

Velocity at this instant v=16.8\ m/s

If the car stops somewhere at a height h

Thus conserving total energy we get

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh

\frac{v^2}{2g}+h_o=h

h=5+\frac{16.8^}{2\times 9.8}

h=5+14.4

h=19.4\ m

6 0
2 years ago
The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is________
maksim [4K]

Answer:

d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

= -∫e²dr/4πε₀r²

= -e²/4πε₀∫dr/r²

= -e²/4πε₀ × -[1/r] from r = +r to -r

W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr

6 0
2 years ago
When you double your speed, it takes about _____ times as much distance to stop?
Zigmanuir [339]
4 times as much distance to stop
8 0
3 years ago
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