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Dennis_Churaev [7]

2 years ago

13

Which of the following can be reduced to a single number in standard form?

1 answer:

raketka [301]2 years ago

6 0

Complete question is;

Which of the following can be reduced to a single number in standard form?

A) 3√3 + 5√8

B) 2√5 + 5√45

C) √7 + √9

D) 4√2 + 3√6

Answer:

Only option B) 2√5 + 5√45 can be reduced to a single number

Explanation:

A) For 3√3 + 5√8;

Let's simplify it to get;

3√3 + 5√(4 × 2)

From this, we get;

3√3 + (5 × 2)√2 = 3√3 + 10√2

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

B) 2√5 + 5√45

Simplifying to get;

2√5 + 5√(9 × 5)

This gives;

2√5 + (5 × 3)√5 = 2√5 + 15√5

Adding the surds gives;

17√5.

This is a single number and thus can be reduced to a single number

C) For √7 + √9

Simplifying, to get;

√7 + 3.

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

D) 4√2 + 3√6

Thus can't be simplified further because both numbers inside the square root don't have factors that are perfect squares.

Thus, it remains 2 numbers and not a single number and can't be reduced to a single number in standard form.

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A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

$p_{rad} = \frac{I}{c}$

<u>Where:</u>

$p_{rad}$ : is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

$p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}$

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th

jek_recluse [69]

Answer:

$a)v_{f}=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}$ (1)

At the beginning the girl is stationary:

$v_{o.girl}=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_{f.girl}=v_{f.ball}=v_{f}$ (3)

We replace (2) and (3) in (1):

$m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\$

We can now solve the equation for v_{f}:

$v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s$

4 0

2 years ago

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where v is the wave's speed and $\lambda$ is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

$f=\frac{400 m/s}{2 m}=200 Hz$

- underwater, the frequency of the wave is:

$f=\frac{1600 m/s}{8 m}=200 Hz$

So, the frequency has not changed.

3 0

2 years ago

Read 2 more answers

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

andreev551 [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

2 years ago