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3 years ago

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A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart

h to Moon, which is 380,000 km?

1 answer:

VladimirAG [237]3 years ago

4 0

Speed = 11000 m/s = 11km/s

D = 380000 km,

t = D/s = 380000 km/ 11km/s

t = 34 545.45 seconds.

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What graph would best represent acceleration as a function of mass when a constant force is applied?

Alex17521 [72]

The graphs of the acceleration as a function of mass and of the acceleration as a function of force are in attachment.

Explanation:

To answer both parts of the question, we refer to Newton's second law, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

a)

To answer this part, we re-arrange the previous equation as follows:

$a=\frac{F}{m}$

Therefore, we notice that if a constant force is applied, the acceleration is inversely proportional to the mass of the object:

$a \propto \frac{1}{m}$

this means that if the mass increases, the acceleration decreases, and if the mass decreases, the acceleration increases.

In a graph of acceleration vs mass, the curve representing this relationship would be a hyperbole. The graph is shown as the first graph in the attached picture.

b)

As before, we re-arrange the previous equation as follows:

$a=\frac{F}{m}$

Here we notice that if the object has a constant mass, the acceleration is directly proportional to the force applied on the object:

$a \propto F$

this means that when the force increases, the acceleration increases, and when the force decreases, the acceleration decreases.

In a graph of acceleration vs force, the curve representing this relationship would be a straight line, as shown in the second graph in the attached picture.

Learn more about Newton's second law of motion here:

brainly.com/question/3820012

#LearnwithBrainly

6 0

3 years ago

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

What mass of electrons would be required to just neutralize the charge of 5.0 g of protons? (the mass of a proton is 1.67262×10−

Step2247 [10]

<span>To neutralize the charge of 1.0g of electrons?</span>

7 0

3 years ago

Read 2 more answers

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel?

SIZIF [17.4K]

-70560, -9.8/2 = -4.9

120^2 = 14400.

Now you multiply 14400 * -4.9 = -70560m.

4 0

4 years ago

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 5.00 T/s. Part A) What is the electric field str

Veronika [31]

Answer:

(A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

Explanation:

Given that,

Magnetic field = 2.0 T

Diameter = 5.0 cm

Rate of decreasing in magnetic field = 5.00 T/s

(A). We need to calculate the electric field strength inside the solenoid at a point on the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

Electric field on the axis of the solenoid

Here, r = 0

$E=\dfrac{0}{2}\times5.00$

$E = 0$

The electric field strength inside the solenoid at a point on the axis is zero.

(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

$E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|$

$E=0.0375= 3.75\times10^{-2}\ V/m$

Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

4 0

4 years ago