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3 years ago

Which of the following quantities can be determined from a speed-time graph of a particle travelling in a straight line?

Physics

2 answers:

enot [183]3 years ago

8 0

The answer is:

Both the distance traveled in a given time and the magnitude of the acceleration at a given instant

Hope I Helped!

Diano4ka-milaya [45]3 years ago

7 0

The answer is letter B

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The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an

blsea [12.9K]

Answer:

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

F = G Me /Re² m

We call gravity acceleration a

g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

R = Re + h

Therefore the attractive force is

F = G Me m / (Re + h)²

Let's take Re's common factor

F = G Me / Re² m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

(1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

F = G Me /Re² m [1- 2 h / Re + 3 (h / Re)²]

F = g₀ m [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

m g =g₀ m [1- 2 h / Re + 3 (h / Re)²]

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

3 0

3 years ago

Astronomy can best be described as a/an

Alchen [17]

<span>Astronomy is a natural science that studies celestial objects and phenomena. More generally, all astronomical phenomena that originate outside Earth's atmosphere are within the purview of astronomy. Therefore, the correct answer to the question "Astronomy can best be described as a/an" is "study of objects beyond the Earth's atmosphere."</span>

3 0

3 years ago

A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi

Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-

iVinArrow [24]

Answer:

B) 3.50 m/s

Explanation:

The linear velocity in a circular motion is defined as:

$v=\omega r(1)$

The angular frequency ( $\omega$ ) is defined as 2π times the frequency and r is the radius, that is, the distance from the center of the circular motion.

$\omega=2\pi f(2)$

Replacing (2) in (1):

$v=2\pi fr$

We have to convert the frequency to Hz:

$7.18rpm*\frac{1Hz}{60rpm}=0.12Hz$

Finally, we calculate how fast is the child moving:

$v=2\pi(0.12Hz)(4.65m)\\v=3.5\frac{m}{s}$

3 0

3 years ago