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3 years ago

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Conduction is the transfer of electrons from a charged object to another object by rubbing. Please select the best answer from t

he choices provided
a. True
b. False

1 answer:

Valentin [98]3 years ago

7 0

The answer to this question is false

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Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

trasher [3.6K]

Answer:

$d_{2}$ = 33.33 cm

Explanation:

Given:

When mass, $m_{1}$ =21 kg

distance travelled is $d_{1}$ = 140 cm

When mass, $m_{2}$ =5 kg

distance travelled is $d_{2}$ = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

$\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}$

$\frac{140}{21}=\frac{d_{2}}{5}$

$d_{2}$ = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

8 0

3 years ago

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o

Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0

3 years ago

How do you calculate speed

Kryger [21]

Answer:

distance(d)=.....

time(t)=....

speed =?

we know that

speed =distance /time

8 0

3 years ago

Read 2 more answers

A horse canters away from its trainer in a straight line, moving 37 m away in 9.1 s. It then turns abruptly and gallops halfway

vesna_86 [32]

Answer:

Average velocity = 1.69 m/s

Average speed = 5.09 m/s

Explanation:

Given that

Horse cover 37 m in 9.1 sec and 18.5 m in 1.8 sec.

As we know that

Average velocity = Displacement / Total time

Average speed = Total distance / Total time

Average velocity:

Average velocity = Displacement / Total time

Total displacement = 37 - 18.5 = 18.5 m

Total time = 9.1 + 1.8 =10.9 s

Average velocity = Displacement / Total time

Average velocity = 18.5 / 10.9

Average velocity = 1.69 m/s

Average speed:

Average speed = Total distance/ Total time

Total distance = 37 + 18.5 = 55.5 m

Total time = 9.1 + 1.8 =10.9 s

Average speed = Total distance/ Total time

Average speed = 55.5 / 10.9

Average speed = 5.09 m/s

3 0

3 years ago

A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g

Arada [10]

Answer:

"Apparent weight during the "plan's turn" is 519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is $\mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}$

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

$a_{n}=\frac{420^{2}}{11000}$

$a_{n}=\frac{176400}{11000}$

$a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}$

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector $W_{\mathrm{app}}=m(\text { vector } g \text { -vector a })$

In magnitude,

$| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}$

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

$\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}$

Which is quite heavier than his/her true weigh of 519.4 N

7 0

3 years ago

Read 2 more answers