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2 years ago

10

A car passes you at 18 m/s to the north and increases its speed at a rate of 3.0 m/s. Determine the car's velocity when it has a

ccelerated for 50 m.

1 answer:

monitta2 years ago

8 0

122

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What tool could help a biologist study the movements of cells? digital camera with a zoom lens a satellite with a digital camera

Annette [7]

I think your answer would be (D) microscope with a video camera
Hope i helped!

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3 years ago

Read 2 more answers

A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100

zavuch27 [327]

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

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2 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Burka [1]

Answer:

$1.91773\times 10^{37}\ kg$

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = $\dfrac{d}{2}=\dfrac{16}{2}=8\ ly$

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$1\ ly=9.461\times 10^{15}\ m$

As the centripetal force balances the gravitational energy we have the following relation

$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg$

Mass of the the massive object at the center of the Milky Way galaxy is $1.91773\times 10^{37}\ kg$

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3 years ago

As the world population continues to rise, the demand for energy also continues to rise. Which type of fuel does little to pollu

Vlad1618 [11]

Answer:

im pretty sure nuclear if not nitrogen

Explanation:

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3 years ago

A helicopter, initially hovering 40 feet above the ground, begins to gain altitude at a rate of 21 feet per second. Which of the

Montano1993 [528]

Answer: y = 40 + 21t

Explanation:

Apply the equation of distance covered.

d = vt + C

Where d is the distance covered

v = velocity , t = time

C = constant = initial distance covered

For the case above....

d = y

y(t) = vt + C

But y(0) =40 = C

C = 40ft

velocity v = 21 ft/s

Therefore, the equation of the altitude is given by;

y(t) = 21t +40

y = 40 + 21t

6 0

3 years ago