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2 years ago

Can any one answer these two

Physics

1 answer:

dalvyx [7]2 years ago

7 0

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

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A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

Sonbull [250]

Answer:

(a) $\overrightarrow{L}=885.5\widehat{k}$

(b) $\overrightarrow{L}=1046.5\widehat{k}$

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, $\overrightarrow{r}=3\widehat{i}-4\widehat{j}$

$\overrightarrow{p}=m\overrightarrow{v}$

$\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )$

$\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}$

So, the angular momentum

$\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=885.5\widehat{k}$

(b) Here, $\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}$

$\overrightarrow{r}=5\widehat{i}-2\widehat{j}$

$\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=1046.5\widehat{k}$

6 0

3 years ago

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

8 0

3 years ago

A child is perched perilously above a 150m ravine. The child has a mass of 45kg. How much

Finger [1]

Answer:

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy is the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8\ m/s^2$ .

The child of mass m=45 Kg is perched above a h=150 m ravine. His gravitational potential energy is:

$U=45~Kg\cdot 9.8\ m/s^2\cdot 150\ m$

U = 66,150 J

6 0

2 years ago

What is the problem with using your boat's engine to drive it onto a trailer?

Elenna [48]

Silt and other wash can accumulate behind the motor and if other boats are added they can get their hulls damaged or scraped then, depending on the size of the material. It can also cause trailers to get stuck if they drive into the place from where the boat came.

3 0

3 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0

3 years ago