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3 years ago

A 30.0 kg box is placed on a 3.00 m tall shelf. What how much work was done on the box?

Physics

1 answer:

Kruka [31]3 years ago

6 0

Answer:

W = 882.9[J]

Explanation:

In order to be able to calculate the work, we must first calculate the force necessary to lift the box. Since the necessary force is equivalent to the weight of the box, we can determine the weight of the box by means of the product of mass by gravitational acceleration.

$w = m*g\\$

where:

w = weight [N]

m = mass = 30 [kg]

g = gravity acceleration = 9.81 [m/s²]

$w = 30*9.81\\w = 294.3 [N]$

Now, the work can be calculated multiplying the force (weight) by the distance [m]

$W = w*d\\W = 294.3*3\\W=882.9[J]$

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a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.

Ivahew [28]

Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?

(7-0)/3.6 = t
t = 1.94 s

6 0

3 years ago

Read 2 more answers

Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay

Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

$w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s$

The maximum velocity is

$v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s$

8 0

3 years ago

Select all of the following that will produce a magnetic field:________.

zysi [14]

Answer:

a ,b , d ,e

Explanation:

took the k12 quiz and got a 100

4 0

2 years ago

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

Montano1993 [528]

Answer:

vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

W = ΔK (1)

for this part the mass is

M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

N -W = 0

N = W = Mg

friction forces have the expression

fr =μ N

fr = μ M g

we substitute in 1

-μ M g x = 0 - ½ M v²

v² = 2 μ g x

let's calculate

v² = 2 0.750 9.8 6.00

v = ra 88.5

v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

p_f = (mₐ + m_b) v

p₀ = p_f

+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

660 vₐ₀ = (660 +490) 9.39 + 490 17.778

vₐ₀ = 19509.72 / 660

vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0

3 years ago

A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat

kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0

3 years ago