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3 years ago

Look at your data. With three foods, Flock v

Physics

1 answer:

wolverine [178]3 years ago

6 0

Answer:the first one was x

the second one is y

Explanation:

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An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

Now, dividing v₂ over v₁, we get the expression:

$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

4 0

3 years ago

The location of the strongest magnetic force is the

Elenna [48]

Answer:

The north pole has the strongest magnetic force

6 0

3 years ago

a whistle you use to call your hunting dog has a frequency of 21 khz, but your dog is ignoring it. you suspect the whistle may n

laila [671]

To solve this problem we will apply the concepts related to the Doppler effect. According to this concept, it is understood as the increase or decrease of the frequency of a sound wave when the source that produces it and the person who captures it move away from each other or approach each other. Mathematically this can be described as

$f = f_0 (\frac{v-v_0}{v})$

Here,

$f_0$ = Original frequency

$v_0$ = Velocity of the observer

$v$ = Velocity of the speed

Our values are,

$v = 340m/s \rightarrow \text{Speed of sound}$

$f = 20kHz \rightarrow \text{Apparent frequency}$

$f_0 = 21kHz \rightarrow \text{Original frequency}$

Using the previous equation,

$f = f_0 (\frac{v-v_0}{v})$

Rearrange to find the velocity of the observer

$v_0 =v (1-\frac{f}{f_0})$

Replacing we have that

$v_0= (340m/s)(1-\frac{20kHz}{21kHz})$

$v_0 = 16.19m/s$

Therefore the velocity of the observer is 16.2m/s

4 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

4 years ago

Pllzzzz help I will mark u as a brainliest

Law Incorporation [45]

Answer:

You dont need BRAINLY there is an app called slader where you scan the textbook barcode and it shows you the answers for the question in the textbook.

Explanation:

8 0

3 years ago