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3 years ago

Can someone plss help I’ll mark u as brainlist!!!

Physics

1 answer:

LekaFEV [45]3 years ago

6 0

Answer:

I believe the answer is A

Explanation:

And i'm sorry if i'm wrong

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A bicycile traves with an average velocty 15km/h north for 20min what is his displasment

Mila [183]

Answer:

5000 m equivalent to 5 Km

Explanation:

Average velocity = $\frac{Displacement}{Time}$

so, Displacement =Average velocity × time

We should convert Km/h to m/s so Km/h ⇒ $\frac{1000}{60*60}$ m/s, also convert time to second so, 20min ⇒(20* 60)seconds

Displacement = (15 × $\frac{1000}{60*60}$ ) × (20×60) =5000m OR 5Km

6 0

3 years ago

The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.

guajiro [1.7K]

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET = pi * d = pi * 1 ft = pi feet

pi feet x 600 rot/min * 60 min /hr * 1 mile / 5280 feet = 21.4 mph

7 0

1 year ago

If an object starts from rest, what is its initial velocity?

Vladimir79 [104]

Answer:

0 m/s

Explanation:

velocity= change in displacement/ time

at rest, the ball does not travel any distance

0/ t

4 0

4 years ago

I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

olga nikolaevna [1]

Answer:

Explanation:

$a = \frac{v_f-v_0}{t}$ which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:

$a=\frac{2.7-1.6}{1.2}$ so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0

3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago