Ask question

Ask question

All categories

2 years ago

13

Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mu

le and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.

1 answer:

MArishka [77]2 years ago

4 0

Answer:

Horizontal component: $F_x = 58\ N$

Vertical component: $F_y = 33.5\ N$

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

$F_x = F * cos(angle)$

$F_x = 67 * cos(30)$

$F_x = 58\ N$

For the vertical component, we have:

$F_y = F * sin(angle)$

$F_y = 67 * sin(30)$

$F_y = 33.5\ N$

So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.

You might be interested in

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0

2 years ago

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0

2 years ago

Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You

AleksAgata [21]

Answer:

$\eta=0.5074\ or\ 50.74\%$

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

density $\rho=1\ g.mL^{-1}$

specific heat $c=4.186\ J.g^{-1}.K^{-1}$

mass of coffee, $m=200\times 1=200\ g$

initial temperature of coffee, $T_i=30^{\circ}C$

final temperature of coffee, $T_f=60^{\circ}C$

power rating of oven, $P=1100\ W$

time taken to reach the final temperature, $t=35\ s$

<u>Heat released by the coffee to come to 60°C:</u>

$Q=m.c.\Delta T$

$Q=200\times 4.186\times 30$

$Q=[tex]\eta=\frac{25116}{49500}$ \ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

$E=P.t$

$E=1100\times 45$

$E=49500\ J$

Now the efficiency:

$\eta=\frac{Q}{E}$

$\eta=0.5074\ or\ 50.74\%$

8 0

2 years ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0

2 years ago

La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción

zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0

2 years ago