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olganol [36]
3 years ago
13

If a bicyclist, with an initial velocity of zero, steadily gained speed until reaching a final velocity of 13m/s, how far would

she travel during this race? (Her time limit is 12s)
Physics
2 answers:
a_sh-v [17]3 years ago
6 0

Answer:

78 meters

Explanation:

v=u+at

13=0+a×12

a=13/12 m/s^2

Now,

v²=u²+2as

13²=2×(13/12)×s

s=13×6=78 meters

Hope, this helps you.

Semmy [17]3 years ago
5 0

Answer:

156m

Explanation:

  • velocity=distance/time
  • V=13, S=?, T=12s

13=S/12

  • cross multiply

S= 13×12

S= 156m

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matrenka [14]
Assuming you are looking for the acceleration a:

1.m_1a = T_1 -m_1g
2.m_2a = m_2g - T_2
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3.\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}
where I = \frac{1}{2} mR^2 and a = \alpha R.

Combining the three equations:
T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }
6 0
3 years ago
If gravity on the earth increased, what affect would it have on the moon
Rufina [12.5K]

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

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3 years ago
Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude
natka813 [3]

Answer:

For vector u, x component = 10.558 and  y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

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Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So u_x=ucos\Theta =16.6\times cos50.5=10.558

Vertical component u_y=usin\Theta =16.6\times sin50.5=12.808

So vector u will be u = 10.558 i+12.808 j

Unit vector u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316

Vertical component v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464

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8 0
3 years ago
Is distance traveled by a moving train qualitative discrete continuous or na?
butalik [34]
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Motion is measured with <em>continuous </em>values; a train might move 300 yards in 1 minute, but we can look at smaller and smaller chunks of time to keep getting shorter and shorter distances. There is no <em />"next" distance the train moves after those 300 yards - it just doesn't make sense for there to be.

It's also measured <em>quantitatively</em>, not <em>qualitatively</em>. This just means that we can use numerical values to measure it, rather than other descriptors like color, smell, or taste.
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Answer:

A, 72 kg•m/s

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6 0
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