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exis [7]
2 years ago
5

A spider accelerates from a standstill to 5m/s in 10s. What is its acceleration?

Physics
1 answer:
weqwewe [10]2 years ago
5 0

Answer:

Acceleration = 0.5m/s²

Explanation:

Given the following data;

Final velocity = 5m/s

Time = 10 seconds

Since the spider started from rest, its initial velocity is equal to 0m/s

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

Substituting into the equation, we have;

a = \frac{5 - 0}{10}

a = \frac{5}{10}

Acceleration = 0.5m/s²

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A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1
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(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

R=\rho \frac{L}{A} (1)

where

\rho is the material resistivity

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A is the cross-sectional area

The radius of the rod is half the diameter: r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m, so the cross-sectional area is

A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega

And now we can re-arrange the eq.(1) to solve for the resistivity:

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(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

where

R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

T=92^{\circ}C

T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

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A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
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Answer:

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Solution:

As per the question:

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