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3 years ago

What charge are electrons

Physics

2 answers:

TEA [102]3 years ago

6 0

Answer: negative

Explanation: it may seem that neutrons would be negative but neutrons are neutral, protons are positive, and electrons are negative

Colt1911 [192]3 years ago

4 0

Answer:

negative

Explanation:

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Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble

Neporo4naja [7]

Answer:

2.4 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v

0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v

0.12 kg m/s = (0.05 kg) v

v = 2.4 m/s

4 0

3 years ago

What happens if magnesium is added to hydrochloric acid (hci) in a single replacement reaction

STatiana [176]

It produces hydrogen gas

3 0

3 years ago

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

$x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s$

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) $\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s$

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0

3 years ago

A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i

lesya [120]

<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

We know:

$\theta_{1}=34\°$

In addition we are told the diffraction grating has 750 slits per mm, this means:

$d=\frac{1mm}{750}$

Solving (2) with the known values we will find $\lambda$ :

$\lambda=(\frac{1mm}{750})sin(34\°)$ (3)

$\lambda=0.00074559mm$ (4)

Knowing $1mm=10^{6}nm$ :

$\lambda=745.59nm$ >>>This is the wavelength of the light, wich corresponds to red.

6 0

3 years ago

Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force

kkurt [141]

Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=

7 0

3 years ago

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