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3 years ago

Unscramble these words TOFO - DNPOU TENOWN - TREEM HINT: they are science words

Physics

2 answers:

padilas [110]3 years ago

8 0

It looks like they are all units of measurement:
FOOT - POUND - NEWTON - METER

Gnoma [55]3 years ago

4 0

FOOT - POUND - NEWTON - METER

These are the correct science terms unscrambled

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A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele

Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0

3 years ago

What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

5 0

3 years ago

0.75 km expressed in centimeters

disa [49]

75000 lol enjoy..............using up 20 characters

7 0

3 years ago

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle

Brilliant_brown [7]

Answer:

$F_2 = 1.10 \mu N$

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

5 0

3 years ago

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

$x=(4.0 m/s)t + (0.85 m/s^3) t^3$

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

$v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2$

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

$v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s$

5 0

3 years ago