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3 years ago

Suppose an object is moving in a straight line at 50 mi/hr. According to Newton's first law of motion, the object will:

Physics

1 answer:

NeX [460]3 years ago

7 0

Answer:

The object will continue to move in straight line at 50 mi/hr except it is acted upon by an external force

Explanation:

Newton First Law of motion: A body will continue in its present state of rest or, if it is in motion will continue to move with uniform speed in a straight line unless it is acted upon by a force. Newton first law of motion is also called the law of inertia.

Inertia: This is the tendency of a body to remain in its states of rest or uniform motion.

From the question, The object moving in a straight line at 50 mi/hr will continue to move in a straight line at 50 mi/hr, except it is acted upon by external force that will change the speed of the object.

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An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet

zimovet [89]

Answer:

F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v²

Em₀ = Emf

e ΔV = ½ m v²

v =√ 2 e ΔV / m

v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

v = √(1.8075 10¹⁶)

v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Emo = U = q DV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v2

Emo = Emf

.e DV = ½ m v2

.v = RA 2 e DV / m

.v = RA (2 1.6 10-19 51400 / 9.1 10-31)

.v = RA (1.8075 10 16)

.v = 1,344 108 m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10-19 1,344 108 0.4

F = 8.6 10-12 N

5 0

3 years ago

Remember to include your data, equation, and work when solving this problem.

andrezito [222]

Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\$

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

$F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]$

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

6 0

3 years ago

Which of these is an example of a mechanical wave?

Oduvanchick [21]

i think the answer is B police siren.

8 0

3 years ago

Read 2 more answers

An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

$f'=f\frac{v\pm v_o}{v\pm v_x}$

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

$v_x=\frac{f}{f'}(v+v_o)-v$

on substituting the values, we get

$v_x=\frac{1790}{1780}(343+2.36)-343$

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0

3 years ago

A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w

Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0

3 years ago