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3 years ago

15

A compressed spring in a toy is used to launch

1 answer:

grandymaker [24]3 years ago

7 0

Kinetic energy = (1/2) (mass) x (speed²)

If the ball was launched in the horizontal direction,
then its total kinetic energy was

   (1/2) x (0.005 kg) x (5 m/s)²

   = (0.0025 x 25) (kg-m²/sec²)

   =   0.0625 Joule .

And that would be the amount of energy stored in the spring.

If the ball was launched at some angle above the horizontal, then
it had some vertical speed too. Its total kinetic energy was then
some greater number, and there was more energy stored in
the spring.

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3 years ago

A wave is produced in a rope. The wave has a speed of 33 m/s and a frequency of 22 Hz. What wavelength is produced? 0. 67 m 0. 7

vekshin1

Answer:

Wavelength = <u>1.5 m</u>

Explanation:

The formula for waves in terms of wavelength, speed and frequency is:

Speed (v) = Frequency (f) × Wavelength (λ)

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3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

A 3 x 10^-6 C charge is 5m away from a -2x 10^-6 C charge A. Attractive because one is positive the other one is negative B. Det

Marizza181 [45]

Answer:

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F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)

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2 years ago