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borishaifa [10]
2 years ago
14

Can someone please answer my question please. thank you ❤️ ​

Physics
1 answer:
faust18 [17]2 years ago
6 0
<h3>₳ ₦ ₷ ₩ € r : </h3>

BATANGAS

  • 560 km

PUERTO PRINCESSA

  • 400 km

DAVAO

  • 480 km

(´。• ω •。`) ♡

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aleksley [76]
The spleen is the mains structure of the lymphatic system! 
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3 years ago
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Need help today please​
Eddi Din [679]

6) True: the car is exerting an equal and opposite force on the truck

7) True: the astronaut can drift back by throwing the tool forward

8) True: the bug exerts an equal and opposite force on you

9) Second Law

Explanation:

6)

This problem can be solved by applying Newton's third law of motion, which states that:

"When an object A exerts a force (action) on another object B, then object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the car and the truck as object A and object B. Here we are told that the truck is exerting a force on the car: therefore, according to Newton's third law, the car is also exerting an equal and opposite force on the truck.

Therefore, the statement is true.

7)

This problem can be also solved by thinking in terms of the Newton's third law of motion.

In fact, at the beginning the astronaut is drifting away from the space station. When he throws away the tool, in the forward direction (away from the space station), he exerts a force on the tool: we can identify this force as the action force, and its direction is away from the space station.

As a result, according to Newton's third law, the tool will also exert a reaction force (equal and opposite) on the astronaut: therefore, the force exerted by the tool on the astronaut is toward the space station, and therefore the astronaut will be pushed back towards the station.

Therefore, the statement is true.

8)

Again, this problem can also be explained using Newton's third law.

In fact, the moment you step on the bug, your foot exerts a force (the action force) on the bug, pushing downward.

As a result, according to Newton's third law, the bug exerts back on you an equal and opposite force (upward). The reason you don't feel this force at all is that your mass is much larger than that of the bug, therefore your acceleration is negligible.

Therefore, the statement is true.

9)

This problem can be explained by using Newton's second law, which states that the net force acting on an object is equal to the product between its mass and its acceleration. Mathematically:

F=ma

where

F is the net force

m is the mass

a is the acceleration

In this problem, the ball slows down as it crosses the field: this means that it has an acceleration (more precisely, a negative acceleration). According to the law, this means also that there is a net, unbalanced force acting on it, in the direction opposite to the motion of the ball. In fact, this force is the force of friction between the ball and the surface.

Learn more about Newton laws of motion:

brainly.com/question/11411375

brainly.com/question/3820012

#LearnwithBrainly

3 0
3 years ago
HELP ASAP
tankabanditka [31]

Answer:

I wanna say 4 I'm not sure though

4 0
2 years ago
A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3
Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

a_c =\dfrac{v^2}{r}

2.32 =\dfrac{3.43^2}{r}

r =\dfrac{3.43^2}{2.32}

   r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 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
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