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2 years ago

13

Is this right? Please help me ITS SOCIOLOGY

1 answer:

Delicious77 [7]2 years ago

7 0

Yes, this is correct Answer.

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HELP!!! When fat comes in contact with sodium hydroxide, it produces soap and glycerin. Determine whether this is a physical cha

dimaraw [331]

Answer:it's a chemical change.physical changes occur without 2 or more reactants,a reaction,and a product

Explanation:

5 0

3 years ago

The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency diffe

mel-nik [20]

Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?

B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?

Answer:

The frequency change percentage is 9.94%

Explanation:

The frequency shift can be calculated as follows.

$F= \frac{V+V_{0}}{V+V_{s}}$

= $9.05GHz\times\frac{343m/s+0}{343m/s+(-31.3m/s)}$

=9.95 GHz

So the frequency change seen by the detector is 9.95 - 9.05

% difference $= \frac{0.9}{9.05} \times100$

= 9.94%

3 0

2 years ago

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

3 years ago

What happens to most of the light waves that strike a clear pane of glass

Marysya12 [62]

It’s C

Because C is a reflection which reflects something such as mirror

Hope this helps! •~•

3 0

2 years ago

Read 2 more answers

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers