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Artemon [7]
3 years ago
13

Что сделать чтоб предмет облодал магнитными свойствами?​

Physics
1 answer:
Brut [27]3 years ago
4 0
What is the question?
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Which best describes the difference between internal and thermal energy?
Dominik [7]
Internal energy cannot be transferred whereas, thermal energy is the energy due to temperature difference
7 0
3 years ago
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A car is accelerating at a rate of 35 m/s2 and has a resulting velocity increase from 17 m/s to 26 m/s. How long did it take to
harina [27]

Answer:The answer is (60 mph - 0 mph) / 8s = (26.8224 m/s - 0 m/s) / 8s = 3.3528 m/s 2 (meters per second squared) average acceleration. That would be 27,000 miles per hour squared.

Explanation:

6 0
3 years ago
A 1200-kg car initially at rest undergoes constant acceleration for 8.8 s, reaching a speed of 10 m/ s. It then collides with a
atroni [7]

Answer:

The final kinetic energy of the two-car system is 60,000 J.

Explanation:

Given;

mass of the car, m = 1200 kg

time of motion, t = 8.8 s

final velocity of the car, v = 10 m/s

Apply the principle of conservation of kinetic energy; the initial kinetic energy is equal final kinetic energy.

K.E_i = K.E_f\\\\K.E_f = \frac{1}{2}mv^2\\\\K.E_f =  \frac{1}{2}(1200)(10)^2\\\\K.E_f = 60,000 \ J

Therefore, the final kinetic energy of the two-car system is 60,000 J.

4 0
3 years ago
g If the momentum of an electron doubles, by what factor would its de Broglie wavelength be multiplied
Fynjy0 [20]

Explanation:

The de broglie wavelength is given by :

\lambda=\dfrac{h}{p}

Here,

h is Planck's constant

p is momentum

Momentum and De-Broglie wavelength has inverse relationship. If momentum of an electron double, its wavelength gets half.

5 0
2 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
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