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4 years ago

6

A shopper in a supermarket pushes a loaded cart with a horizontal force of 11 N. The cart has a mass of 35 kg. The acceleration

of gravity is 9.8 m/s 2 . Disregarding friction, how far will the cart move in 3.5 s, starting from rest? Answer in units of m.

1 answer:

Tcecarenko [31]4 years ago

7 0

Answer:

s = 1.925 meters

Explanation:

Given that,

Horizontal force acting on the cart, F = 11 N

Mass of the cart, m = 35 kg

To find,

Distance covered by the cart in 3.5 seconds if it starts from rest.

Solution,

As the cart is starting from rest, u = 0

Let x is the distance moved by the cart. It can be calculated using the second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Since, $a=\dfrac{F}{m}$

$s=\dfrac{1}{2}\times \dfrac{F}{m}\times t^2$

$s=\dfrac{1}{2}\times \dfrac{11}{35}\times (3.5)^2$

s = 1.925 meters

Therefore, the distance covered by the cart is 1.925 meters.

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Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

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Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

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3 years ago

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

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Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

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3 years ago

This is “Fusion Reactions”.<br> Please answer number 8. Thank you.

Angelina_Jolie [31]

Answer:

²₁H + ³₂He —> ⁴₂He + ¹₁H

Explanation:

From the question given above,

²₁H + ³₂He —> __ + ¹₁H

Let ⁿₐX be the unknown.

Thus the equation becomes:

²₁H + ³₂He —> ⁿₐX + ¹₁H

We shall determine, n, a and X. This can be obtained as follow:

For n:

2 + 3 = n + 1

5 = n + 1

Collect like terms

n = 5 – 1

n = 4

For a:

1 + 2 = a + 1

3 = a + 1

Collect like terms

a = 3 – 1

a = 2

For X:

n = 4

a = 2

X =?

ⁿₐX => ⁴₂X => ⁴₂He

Thus, the balanced equation is

²₁H + ³₂He —> ⁴₂He + ¹₁H

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3 years ago

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Firstly we need to know this velocity has two components:

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$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

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Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

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