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4 years ago

What objects were formed first, second, etc. after the Big Bang?

1 answer:

3 0

The first objects ever made were dust particles smaller than the molecule. They then formed the molecule which then formed the atom.

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A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the

REY [17]

Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

4 0

4 years ago

9. A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particu

kotykmax [81]

Voltage =Current * Resistance

= 9 * 43

Voltage = 387 V

6 0

3 years ago

Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.

Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1 : m1= 5kg

50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

F2 = 2*a

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0

3 years ago

If the speed of an object is increasing, then the forces acting on that object must be ________.

In-s [12.5K]

Answer:

Unbalanced.

Explanation:

Usually, unbalanced forces cause acceleration, or increased movement.

5 0

3 years ago

Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

disa [49]

Answer:

The net magnetic field ta the center of square is $1.36\times10^{-4} T$ .

Explanation:

Current, I = 12 A , side ,a = 10 cm = 0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

$B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 + sin 45 \right )\\\\B = 3.4\times 10^{-5} T$

Net magnetic field at the center of the square is

B' = 4 B

$B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T$

4 0

3 years ago