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3 years ago

7

La resultante de dos fuerzas perpendiculares aplicadas a un mismo cuerpo es 11.18 N y el módulo de una de ellas es de 10 N. ¿Cuá

l es el módulo de la otra?

1 answer:

Dmitrij [34]3 years ago

3 0

Answer:

English?

Explanation:

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Solve for the unknown in each of these circuits

Tamiku [17]

<u> Ohms law: </u> This law relates voltage difference between two points. Mathematically, the law states that V=IR;

Where

V = voltage difference ; in volts

I = Current ; in Amperes

R = Resistance ; in ohms

<u>1. Answer : </u> given that R = 10 ; V= 12 V ; I = ?

From ohms law, I = V/R

= 12/10

= 1.2 Amp.

<u>2. Answer:</u> given that R = 10 ; V= ? ; I = 5

From ohms law, V = IR

= 10×5 = 50 V

<u>3 . Answer:</u> given that R = ? ; V= 120 ; I = 5

From ohms law, R = V/I

= 120/5

= 24 Ω

<u>4 . Answer:</u> given that R = ? ; V= 10 ; I = 20

From ohms law, R = V/I

= 10/20

= 0.5 Ω

<u>5 . Answer:</u> given that R = 480 ; V= 24 ; I = ?

From ohms law, I = V/R

= 24/480

= 0.05 A

<u>6. Answer:</u> given that R = 150 ; V= ? ; I = 1

From ohms law, V = IR

= 1 × 150

= 150 V

6 0

4 years ago

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass <u><em>m</em></u> , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

4 years ago

Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

2 years ago

Which ball moved at the same speed as Ball 3?

Luden [163]

Answer:

you forgot to attach the image

3 0

3 years ago

Read 2 more answers