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Nikolay [14]
3 years ago
13

Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture

Jerry using a net dropped from an bridge that is 15m high, how much time before Jerry is under the bridge, should Tom drop the net? s How far away from the bridge is Jerry when Tom drops the net? m
Physics
1 answer:
Kruka [31]3 years ago
3 0

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

v_{0y}: is the initial vertical velocity of the net = 0 (it is dropped from rest)

0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}

t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

v = \frac{x}{t}

x = v*t = 18 m/s*1.75 m = 31.5 m

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!                                                                    

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irga5000 [103]

Answer:

1.5048\times 10^{-23}\ Am^2

Explanation:

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r = Radius of circle = 5.7\times 10^{-11}\ m

v = Velocity of proton = 3.3\times 10^6\ m/s

Magnetic moment is given by

M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2

The magnetic moment associated with this motion is 1.5048\times 10^{-23}\ Am^2

5 0
3 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte
timurjin [86]

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

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nika2105 [10]

Answer:

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Explanation:

Given data

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That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

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0.5*35*100=981h

1750=981h

h= 1750/981

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Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

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3 years ago
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