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lesya [120]
3 years ago
5

If Manuel exerts a force of 10 N to push a desk to the right at the same time Lynn exerts a force of 15 N to push the desk to th

e left, the desk will move to the
left
right
sideways
it won't move
Physics
1 answer:
wolverine [178]3 years ago
3 0

Answer:

The desk will move to the left.

Explanation:

Since Manuel is exerting a force of 10 N to the right, and Lynn is exerting a force of 15 N to the left, it means that Lynn is exerting a force of 5 N more than Manuel, meaning that the desk will move to the left, the direction Lynn is pushing it too.

<em>Please </em><em>like </em><em>and </em><em>mark </em><em>brainliest</em><em>!</em>

<em>Hope </em><em>it </em><em>helps</em><em>!</em>

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A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th
Tamiku [17]

Answer:

461.88 N

Explanation:

F_{g} = Weight of the swing = 800 N

T = Tension force in the rope

F = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N

Using equilibrium of force in horizontal direction using the force diagram, we get

F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N

6 0
3 years ago
Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard
podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0
3 years ago
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv
PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The  largest total area of the oil slick  A = 8.257 *10^{9} \ m^2

Explanation:

From the question we are told that

     The volume of oil the escaped is  V  = 18000 \ L

    The refractive index of oil is n_o = 1.1

     The refractive index of water is n_w = 1.33

      The wavelength of the light  is \lambda = 485 \ nm =  485 * 10^{-9} \ m

         

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

          d = m *\frac{\lambda}{2n_w}

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

      d = 1 * \frac{485 *10^{-9}}{2 * 1.1}    

      d = 218 nm    

The are can be mathematically evaluated as

        A = \frac{V}{d}

Substituting values

        A = \frac{18000}{218*10^{-8}}

        A = 8.257 *10^{9} \ m^2

6 0
3 years ago
Pleaseeee Please help, I will love you forever and ever
matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

6 0
3 years ago
.hjvhhchhvvgvfggghhjidaaawryui<br>​
Naddik [55]

Answer:

yes

Explanation:

i do agree

5 0
3 years ago
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