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zzz [600]
3 years ago
14

Need help with is if u can asap

Physics
1 answer:
jeka943 years ago
8 0
I am guessing by what I had it is a plz don't quote me on that I could be wrong
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Which method of testing substances is a sure way to identify a chemical reaction?
Sergeu [11.5K]

Answer:

B

Explanation:

That's the answer. Hope it helped!

7 0
2 years ago
Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3​
kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0
3 years ago
An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts
Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0
3 years ago
Read 2 more answers
Trick question (really easy) just for fun! If u get this right u get brainliest!
Maurinko [17]

Answer:

piper duh

Explanation:

6 0
3 years ago
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A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be
ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

                                      = 3700\times \dfrac{2\pi}{60}=387.27\ rad/s

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}

\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}

\alpha = -259.28 rad/s^2

3 0
3 years ago
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