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gladu [14]
3 years ago
12

A tiger paces back and forth along the front of its cage, which is 8 m wide. The tiger starts from the right side of the cage, p

aces to the left side, then back to the right side, and finally back to the left. What total distance has the tiger paced? What is the tiger's resultant displacement?
Physics
1 answer:
Black_prince [1.1K]3 years ago
3 0

Answer:

The total distance is 24 m.

The tiger's resultant displacement is 8 m.

Explanation:

Given that,

Width of tiger= 8 m

Tiger starts from the right side of the cage,

Paces to the left side

\Delta x_{1}=+8\ m

Then, back to the right side,

\Delta x_{2}=-8\ m

finally, back to the left

\Delta x_{3}=+8\ m

We need to calculate the total distance covered

Using formula of distance

d= x_{1}+x_{2}+x_{3}

Put the value into the formula

d=8+8+8

d=24\ m

We need to calculate the tiger's resultant displacement

Using formula of displacement

D=\Delta x_{1}+\Delta x_{2}+\Delta x_{3}

Put the value into the formula

D=8+(-8)+8

D=8\ m

Hence, The total distance is 24 m.

The tiger's resultant displacement is 8 m.

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Answer:

-963.93 m/s²

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s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
3 years ago
A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the s
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Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

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Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

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Thus,

\alpha=\frac {\Delta r}{r\times \Delta T}

Also,

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So,

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Suppose a straight wire with a length of 2.0 m runs perpendicular to a magnetic field with a magnitude of 38 T. What current wou
iragen [17]

Answer:

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The weight of a student with a mass of m = 75 kg is:

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We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is

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B = 38 T is the magnetic field

\theta=90^{\circ} is the angle between the direction of B and L

Since we want W=F, we can write

ILB sin \theta=W

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8 0
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