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2 years ago

Compute the binomial expansion for (1+x)^5

Physics

1 answer:

blondinia [14]2 years ago

6 0

Answer:

1+5x+10x^2+10x^3+5x^4+x^5

Explanation:

You just make it (1+x)(1+x)(1+x)(1+x)(1+x) and multiply it out until it's all one big term.

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A 36-cm-circumference loop of wire is placed between the poles of an electromagnet. When a field of 0.76 T is switched on in a t

Marrrta [24]

Answer:

R = 25.3ohms

Explanation:

See attachment below.

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2 years ago

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Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

$ma = \text{Mass}\times \text{Acceleration}$

$ma = kg \cdot \frac{m}{s^2}$

At the same time we have that

$\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2$

$\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2$

$\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}$

Therefore there is not have same units and both are not consistent and the correct answer is B.

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3 years ago

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

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2 years ago

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if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.

Likurg_2 [28]

Answer:

4 seconds

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -5 m/s²

Find: t

v = at + v₀

0 m/s = (-5 m/s²) t + 20 m/s

t = 4 s

6 0

3 years ago

A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes

Lerok [7]

Answer:

R = 715.4 N

L = 166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L = 166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

0 = 0 checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

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2 years ago