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2 years ago

Hannah drove 360 mi in 5.2 hours. What was her average speed?

1 answer:

5 0

Her average speed would be 69.2 miles per hour

Explanation:
When you take the number of miles and divide it by the number of hours you get 69.2 (when rounded)
That is how fast she went per hour

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Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s

PSYCHO15rus [73]

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a , a is separation of slits

Puting the values given

.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

8 0

2 years ago

A 0.280 m radius, 475 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

KonstantinChe [14]

Hope it helps u mate. Plz follow me .

8 0

3 years ago

A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction

Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

$v=\frac{10}{2}=5m/s\;\cdots(i)$

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, $d=vt$

From equation (i)

$\Rightarrow d=5t\;\cdots(ii)$

As the jogger starts from origin, so, the distance, $d$ , also represents the position of the jogger at the time $t$ s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

$d=5\times 5=25 m$

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

$d=5\times 15=75 m$

So, the jogger is at a distance of 75 m from the origin.

8 0

2 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$