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3 years ago

What is the net force on the box?

Physics

1 answer:

omeli [17]3 years ago

5 0

Answer:

The magnitude of the force net is an acting object multiplied by acceleration of the object

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This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s

Dennis_Churaev [7]

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

5 0

3 years ago

Eleven wa=eighs 47 kg. her height is 1.63 m. what is her bmi

valentinak56 [21]

Answer:

17.7kg/ $m^{2}$

Explanation:

BMI =

= $\frac{weight (kg)}{height (m) . height (m)}$

= 47kg/(1.63m×1.63m)

= 17.689kg/ $m^{2}$

≈ 17.7kg/ $m^{2}$

6 0

3 years ago

N Rutherford's gold foil experiment the alpha particles pass through which part of the atom?

OleMash [197]

It passes through C the Electron Cloud

3 0

3 years ago

Please help on this one?

pashok25 [27]

Answer:

B) It is connected in parallel and measures potential differences.

Explanation:

A voltmeter measures electrical potential differences between two points in an electric circuit. It is also connected in parallel to measure a device's voltage, while an ammeter is connected in series to measure a device's current.

Hope this helps! :)

7 0

3 years ago

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 180-turn coil that has

OlgaM077 [116]

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

$\omega=2\pi f$

Plug in the value of 'f' and solve for 'ω'. This gives,

$\omega=2\pi\times 60.0=120\pi\ rad/s$

Now, the maximum emf of the generator is given by the formula:

$E=NAB\omega$

Rewriting in terms of magnetic field, 'B', we get:

$B=\frac{E}{NA \omega}$

Plug in the given values and solve for 'B'. This gives,

$B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T$

Therefore, the magnitude of the magnetic field in which the coil rotates is 0.082 T.

8 0

4 years ago