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3 years ago

10

5. The Weeks family took a trip to the Dallas 200 for

1 answer:

zheka24 [161]3 years ago

4 0

Answer: idk

Explanation: hahahaha

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Microwelds are formed where

Jlenok [28]

Between the bumps and dips of two surfaces. SO the answer is 2 surfaces. Hope this helps! :)

8 0

3 years ago

Read 2 more answers

A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta

fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$

4 0

3 years ago

Graph are pictorical representations of

inn [45]

Answer:

Graphs are pictorial representations of relationships.

6 0

3 years ago

Read 2 more answers

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving

Papessa [141]

Answer:

The final velocity of the object after 2 seconds is 30 m/s

Explanation:

Given;

constant downward acceleration, a = 10 m/s²

initial velocity of the object falling down, v = 10 m/s

time of fall, t = 2 s

The final velocity of the object is given by;

v = u + at

where;

v is the final velocity

v = 10 + (10)(2)

v = 10 + 20

v = 30 m/s

Therefore, the final velocity of the object after 2 seconds is 30 m/s

3 0

3 years ago