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Ierofanga [76]
3 years ago
12

An electron has a mass of 9.1x10-31 kg. What is

Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

3.185×10^-29 kgm/s

Explanation:

Momentum(p)=mass×velocity

=9.1×10^-31×3.5×10

=3.185×10^-29 kgm/s

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What is the density of a piece of wood with a mass of 25 kg<br> and a volume of 0.0385 m³?
Orlov [11]

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

p =  \frac{mass \: (in \: kg)}{volume \: (in \:  {m}^{3}) }

We can substitute the givenmass and volume to find density of the object.

p =  \frac{25kg}{0.0385 {m}^{3} }  \\  = 649kg \: per \:  {m}^{3}

Therefore the density of this object is 649kg/m^3.

7 0
2 years ago
A series circuit is composed of two resistors:
WITCHER [35]
<h3>Answer:</h3>

1.3 Amps

<h3>Explanation:</h3>

<u>We are given;</u>

A circuit with resistors, R1 and R2

R1 = 7 Ω

R2 = 11 Ω

Voltage = 24 V

We are required to calculate the current in the circuit.

<h3>Step 1: We need to find the effective resistance.</h3>

When resistors are arranged in series, the effective resistance is calculated by;

Rt = R₁ + R₂ + R₃ + ..........Rₙ

Therefore;

Total resistance = 7 + 11

                          = 18 Ω

<h3>Step 2: Calculate the current in the circuit</h3>

From the ohm's law;

V = IR

Rearranging the formula;

I = V/R

Thus;

I = 24 V ÷ 18 Ω

 = 1.333 Amps

= 1.3 Amps

Thus, the current in the circuit is 1.3 Amps

7 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr
AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

8 0
3 years ago
Why is a shadow formed
a_sh-v [17]

Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

6 0
3 years ago
Read 2 more answers
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