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2 years ago

The law of inertia applies to objects

Physics

2 answers:

Anestetic [448]2 years ago

6 0

Answer:

at rest and in motion

Explanation:

The law of inertia applies to objects at rest and in motion

abruzzese [7]2 years ago

6 0

Answer:

Explanation: At rest and in motion because an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

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Conductivity in a metal results from the metal atoms having

maks197457 [2]

Answer: A) highly mobile electrons in the valence shell

Explanation: conductivity in metals is a result of the movement of electrically charged particles—the electrons. These free electrons also known as valence electrons are free to move, and as a result they can travel through the lattice that forms the physical structure of a metal. The presence of valence electrons determines a metal's conductivity. However, several other factors can affect the conductivity of a metal such as impurities, temperature, magnetic fields etc.

5 0

3 years ago

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

8 0

2 years ago

A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

T = F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm $= \frac{2}{100} = 0.02m$ for z the perpendicular distance

So Elbow Torque is $T_e= 4000 * 0.02$

$= 80Nm$

To obtain the torque required we substitute 300 N for F and 30cm $=\frac{30}{100} = 0.3 m$

So the Required Torque is $T_R = 300 *0.3$

$=90Nm$

Now since $T_e < T_R$ it mean that the weight lifter would not get past this sticking point

7 0

2 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 18 cm that is placed in a spatially uniform m

kozerog [31]

Answer:

ε = 6.617 V

Explanation:

We are given;

Number of turns; N = 40 turns

Diameter;D = 18cm = 0.18m

magnetic field; B = 0.65 T

Time;t = 0.1 s

The formula for the induced electric field(E.M.F) is given by;

ε = |-NAB/t|

A is area

ε is induced electric field

While N,B and t remain as earlier described.

Area = π(d²/4) = π(0.18²/4) = 0.02545

Thus;

ε = |-40 × 0.02545 × 0.65/0.1|

ε = 6.617 V

(we ignore the negative sign because we have to take the absolute value)

6 0

2 years ago

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

2 years ago