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3 years ago

A car accelerates uniformly from rest and

Physics

1 answer:

miv72 [106K]3 years ago

7 0

Answer:

29.75 revolutions

Explanation:

The kinematic formula for distance, given a uniform acceleration a and an initial velocity v₀, is

$d=v_0t+\frac{1}{2}at^2$

This car is starting from rest, so v₀ = 0 m/s. Additionally, we have a = 9.2/9.7 m/s² and t = 9.7 s. Plugging these values into our equation:

$d=0t+\frac{1}{2}\left(\frac{9.2}{9.7}\right)(9.7)^2\\d=\frac{1}{2}(9.2)(9.7)\\d=4.6(9.7)\\d=44.62$

So, the car has travelled 44.62 m in 9.7 seconds - we want to know how many of the tire's circumferences fit into that distance, so we'll first have to calculate that circumference. The formula for the circumference of a circle given its diameter is $c=\pi{d}$ , which in this case is 47.8π cm, or, using π ≈ 3.14, 47.8(3.14) = 150.092 cm.

Before we divide the distance travelled by the circumference, we need to make sure we're using the same units. 1 m = 100 cm, so 105.092 cm ≈ 1.5 m. Dividing 44.62 m by this value, we find the number of revs is

$44.62/1.5\approx29.75$ revolutions

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A ball is kicked at 30.0 m/s at an angle of 20.0°. Beneath the tilted columns calculate the vertical and horizontal components.

RSB [31]

The velocities at time t are

• Horizontal:

v = (30.0 m/s) cos(20.0º)

• Vertical:

v = (30.0 m/s) sin(20.0º) - g t

(where g = 9.80 m/s² is the magnitude of the acceleration due to gravity)

If you only want the initial velocities, they are

• Horizontal:

v = (30.0 m/s) cos(20.0º) ≈ 28.2 m/s

• Vertical:

v = (30.0 m/s) sin(20.0º) ≈ 10.3 m/s

(just set t = 0)

As far as starting equations go, you can derive everything from the definition for average acceleration:

a = ∆v / ∆t = (final v - initial v) / t

→ v = u + a t

(here, u stands in for "initial v" and v is simply velocity at time t )

There is no acceleration in the horizontal direction, while the ball is essentially in free-fall in the vertical direction.

6 0

3 years ago

If the ice cube absorbs heat, then, heat

Rashid [163]

Answer:

<h2>I have no idea what any of that means :'(</h2>

Explanation:

<h2>So Sorry. ΓΟΛ3</h2>

5 0

3 years ago

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by

Artemon [7]

Answer:

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

$E=I^2Rt$ . ....(1)

As,

$I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A$

As per Ohm's law, V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega$

Using formula (1).

$E=0.835^2\times 14.37\times 2\\\\=20.03\ J$

So, the energy delivered to the lightbulb filament is 20.03 J.

6 0

3 years ago

(a) calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the radius of the earth is

kakasveta [241]

(a) The value of the gravitational acceleration is given by:

$g=\frac{GM}{r^2}$

where

$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

M is the Earth's mass

$r=6371 km=6.371 \cdot 10^6 m$ is the Earth's radius at the pole

If we use the value for g given by the problem, $g=9.830 m/s^2$ , and we rearrange the equation above, we find the value of the Earth's mass:

$M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg$

(b) The value we found for the Earth's mass is $5.982 \cdot 10^{24} kg$ , and we see that this value is slightly larger than the accepted value of $5.979 \cdot 10^{24} kg$ .

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3 years ago

5. YOU SHOULD USUALLY DRIVE SLOWER

saveliy_v [14]

Answer:

The answer is C.

Explanation:

If there are brake lights then you should drive slower.

3 0

3 years ago

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