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3 years ago

8

A biker first accelerates from 0.0 m/s to 6.0 m/s in 6 s, then continues at this speed for 5 s. What is the total distance trave

led by the biker?
100 m

80 m

30 m

48 m

Answer and I will give you brainiliest
PLZ heeeeelp

((((WITH STEPS))))

1 answer:

kompoz [17]3 years ago

4 0

Answer:

100m

Explanation:

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An object is dropped from a height of 25 meters. At what velocity will it hit the ground?

Goryan [66]

You can use Vf^2-Vi^2 = 2ax

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Or you can use energy

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A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

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An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

andrew-mc [135]

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

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Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

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Furkat [3]

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