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4 years ago

If an object with the action of a force, is done.

Physics

1 answer:

elena-s [515]4 years ago

3 0

Work is done when a force causes an object to move therefore c. moves, work

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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca

gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B) y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D) B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

F = ma

q E = m a

a = q / m E

We use the kinematic relationship

y = voy t - ½ to t²

x = v₀ₓ t

The initial vertical velocity of electrons is zero

y = ½ a (x / v₀ₓ)²

We replace

y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

Fm = Fe

q v B = q E

v = E / B

Part D

We replace

y = ½ (q / m) E L² / (E / B)²

y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

-4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

-16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

B² = -2.59 10⁻¹ q / m

q / m = -1.758 10¹¹ C/ kg

B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0

3 years ago

If the coefficient of kinetic friction between a crate and the floor is 0.20,

kow [346]

Answer:

0.3 newtons

Explanation:

4 0

3 years ago

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

⠀

$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

⠀

The horizontal range of projectile at x.

⠀

$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

⠀

$\large\textsf{\underline{Now substituting the required values}}$

⠀

$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

⠀

The vertical position of projectile at y.

⠀

$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

⠀

$\textsf{ \large {\underline{Now substituting the required values}} }$

⠀

$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$