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olga nikolaevna [1]

4 years ago

13

Question 9 please. How do you find instantaneous velocity and how do I sketch a velocity vs time graph from a position vs time o

ne?

1 answer:

pickupchik [31]4 years ago

3 0

How do you find instantaneous velocity

Select a point on a distance-time curve graph. Draw a tangent to the curve at that point. Tangent -> hypotenuse of right angled triangle. Opp/adjacent in graph units is vel at that point -> in distance and/or time

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WILL GIVE BRAINLIEST DON'T KNOW IF I SPELLED IT WRONG BUT U KNOW WHAT I MEAN:)

eimsori [14]

Answer: well jabberwocky means like happy or like good vs evil

Synonyms: abracadabra, double talk, gibberish, gobbledygook, mumbo jumbo, babble, blather, blatherskite, gabble, jabber and jargon

Definition: invented or meaningless language; nonsense.

Explanation:

3 0

3 years ago

Read 2 more answers

A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

$0= u^2-2as_{1}$

$s_{1}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

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

Now, put the value of a in equation (I)

$s_{1}=\dfrac{8}{F}$

Now, For second object

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$0= u^2-2as_{2}$

$s_{2}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$F = ma$

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

Now, put the value of a in equation (I)

$s_{2}=\dfrac{8}{F}$

Hence, Both objects travel the same distance.

6 0

3 years ago

The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and

mina [271]

Answer:

$\frac{dQ}{dt}= 4312 W$

Explanation:

As we know that base of the slab is given as

$A = 11 \times 8$

$A = 88 m^2$

now we know that rate of heat transfer is given as

$\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)$

here we know that

$k = 1.4 W/m k$

Also we have

$x =0.20$

$\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)$

$\frac{dQ}{dt}= 4312 W$

7 0

3 years ago

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

$V = \frac{4}{3}\pi R^3$

Replacing at the equation we have that

$\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}$

$\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}$

$\rho = 5.75*10^{17}kg/m^3$

Therefore the density of a neutron star is $5.75*10^{17}kg/m^3$

4 0

3 years ago

If you are driving at a constant velocity of 5 m/s what is your<br> acceleration?

Paraphin [41]

"Acceleration" means the rate at which velocity is changing.

If velocity is constant, then it isn't changing.

If velocity isn't changing, then acceleration is <em>zero</em>.

3 0

3 years ago