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Reil [10]
3 years ago
14

Answer plz also give reason

Physics
1 answer:
user100 [1]3 years ago
4 0

See this suggested solution.

1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.

2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.

3. using the conditions described in 2. the answer is C.

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You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30°C to
earnstyle [38]

The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.

We are asked to calculate the velocity of sound at 36 degree celsius.

Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.

The velocity at this temperature is calculated as -

                            V = 331  + 0.6T m/s

                               = 331 + 0.6 × 36 m/s

                               = 331 + 21.6 m/s

                               = 352.6 m/s.

Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

5 0
3 years ago
Read 2 more answers
Debido al desorden en el laboratorio un científico tiene 2 termómetros diferentes pero no sabe en qué escalas están por lo que d
just olya [345]

Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

F=(\frac{9}{5})C+32 (I)

Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos C=18 en la ecuación (I) deberíamos obtener F=64.4 ⇒

F=(\frac{9}{5}).(18)+32=32.4+32=64.4

Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.

6 0
2 years ago
Which inference about Arthur is supported by the text?
yarga [219]
There is no picture
8 0
3 years ago
For high and low tides differences would they be caused by the moon and how?
Sunny_sXe [5.5K]
The moon has a small amount of gravity. Low tides mean the moon is not pulling on the water. High tides mean that the moon is pulling on the water.
8 0
2 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
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