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2 years ago

8

A biker accelerates from 22m/s to 62m/s. His acceleration was 10m/s2. How long did it take to speed up? (select the correct equa

tion)

1 answer:

xenn [34]2 years ago

5 0

Answer:

Option 4

Explanation:

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In the two-slit experiment, for the condition of bright fringes, the value of m = +2 corresponds to a path difference of λ.

melisa1 [442]

Answer:

False

Explanation:

We know that if path difference is even multiple of wavelength then bright fringes are formed and if path difference is odd multiple of wavelength then dark fringes are formed .

For bright fringes

Path difference Δx = m λ

m = 0 , 2 , 4 , 6.......

If m = 2 then the path difference will be

Δx = 2 λ

therefore the above statement if false.

False

7 0

3 years ago

A bodybuilder deadlifts a 215 kg weight to a height of 0.90 m above the ground. If he deadlifts this weight 10 times in a span o

wariber [46]

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

<h3>What is power?</h3>

In physics, power (P) is the work (W) done over a period of time.

Step 1. Calculate the work done by the bodybuilder each time.

The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.

W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N

Step 2. Calculate the work done by the bodybuilder over 10 times.

W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N

Step 3. Calculate the power exerted by the bodybuilder.

The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.

P = 1.9 × 10⁴ N/45 s = 421 W

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

Learn more about power here: brainly.com/question/911620

#SPJ1

4 0

2 years ago

A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T

Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

a) The minimum force of the wind that turns the system is Fw = 17.64 N

b) The system resists much greater forces because the base has more mass

Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

Σ τ = 0

Where τ is the torque

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

cos 15 = $\frac{y_w}{2.35}$

$y_w$ = 2.35 cos 15

Post Weight

sin 15 = $\frac{x_p}{2.00}$

xp = 2.0 sin 15

Base weight

cos (90-15) = $\frac{x_b}{0.25}$

xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

$F_w \ y_w + W_p \ x_p - W_b \ x_b = 0$

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is $W_p$ = 26.0 N and the weight of the base with water is $W_b$ = 810 N

$F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}$

$F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}$

Let's calculate

$F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}$

$F_w$ = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

a) the minimum force of the wind that turns the system is Fw = 17.64 N

b) the system resists much greater forces because the base has more mass

Learn more here: brainly.com/question/7031958

3 0

3 years ago

A jewellery shop owner has two identical clear gemstones but cannot remember which one is diamond and which one is rutile. Rutil

pantera1 [17]

We know

$\boxed{\sf n_{21}=\dfrac{C}{V}}$

How he find:-

The owner will measure speed of light through the index(Diamond or rutile).Then using calculations he fill find which is the velocity of diamond or rutile

<h3> For diamond</h3>

$\\ \sf\longmapsto n_D=\dfrac{C}{V_D}$

$\\ \sf\longmapsto 2.4=\dfrac{3\times 10^8ms^{-1}}{V_D}$

$\\ \sf\longmapsto V_D=\dfrac{3\times 10^8ms^{-1}}{2.4}$

$\\ \sf\longmapsto V_D=1.25\times 10^8ms^{-1}$

$\\ \sf\longmapsto V_D=125\times 10^6ms^{-1}$

<h3>For rutile</h3>

$\\ \sf\longmapsto n_R=\dfrac{C}{V_R}$

$\\ \sf\longmapsto V_R=\dfrac{C}{n_R}$

$\\ \sf\longmapsto V_R=\dfrac{3\times 10^8ms^{-1}}{2.9}$

$\\ \sf\longmapsto V_R=1.03\times 10^8$

$\\ \sf\longmapsto V_R=103\times 10^6ms^{-1}$

5 0

3 years ago

In the qualifying round of the 50-yd freestyle in the sectional swimming championship, Dugan got an early lead by finishing the

Nata [24]

Answer:

v = 0

Explanation:

Given that,

Total distance is 50 yards

Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds

Dugan finished the return leg (25.00 yd distance) in 10.22 seconds.

We need to find Dugan's average velocity for the entire race. As he returns at the initial position. As a result, the net displacement is equal to 0. So,

Average velocity = net displacement/time

v = 0

Hence, his average velocity for the entire race is 0.

6 0

4 years ago