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4 years ago

6

For the following distance vs time graph

1 answer:

Setler [38]4 years ago

6 0

Answer:

3

5

4

Explanation:

x = (8=(9+9)

(9+9) = 4, 5, 3

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Calculate the speed of an object that moves 200 m in 4 s?

Rasek [7]

50m/s

I divided the 200m by 4s (so 200/4) to get 50m per second

7 0

3 years ago

A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc

Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g= $\frac{15}{1000}=0.015 kg$

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

$\frac{1}{2}(m+M)^2V^2=(m+M)gh$

Using g= $9.8m/s^2$

Substitute the values

$\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086$

$V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}$

$V=\sqrt{2\times 3.015\times 9.8\times 0.086}}$

$V=1.3m/s$

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

$mv+Mv'=(m+M)V$

Using the formula

$0.015v+3(0)=3.015(1.3)$

$0.015v=3.015(1.3)$

$v^2=\frac{3.015(1.3)}{0.015}=261.3$

$v=261.3 m/s$

5 0

4 years ago

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

Will give brainliest to right answer!

viva [34]

The correct answer is D.

8 0

4 years ago

What is one disadvantage of a series circuit?

nlexa [21]

If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺

6 0

4 years ago