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Ksivusya [100]
1 year ago
9

To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m

. If the work done by the student is 32 J, what is the magnitude of the force he exerts?
Physics
1 answer:
hoa [83]1 year ago
3 0

Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

F= ?

W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

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Una columna de mármol, cuya área de sección transversal es de 2.0 m2 sostiene una masa de 25.000 kg. Encontrar: (3 pto )a) El es
bazaltina [42]

Responder:

122,500 Pa; 2.45 × 10 ^ -6; 2.94 × 10 ^ -5m

Explicación:

Dado lo siguiente:

Área de sección transversal (A) = 2m ^ 2

Masa (m) = 25000 kg

Módulo de Young = 50 x 10 ^ 9 N / m2

(1) estrés en la columna:

Estrés = Fuerza / Área

F = masa * aceleración debido a la gravedad

F = 25000kg * 9.8m / s ^ 2 = 245,000J

Estrés = 245,000J / 2m ^ 2

Estrés = 122,500 Pa

2) Deformación de la unidad (deformación):

Usando la relación:

Módulo de Young = Estrés / tensión

50 × 10 ^ 9 = 122,500 / CEPA

Cepa = 122500 / (50 × 10^9)

Cepa = 0.00000245

C) Si la altura es de 12 m, ¿cuánto se acorta la columna?

Cepa = extensión / longitud

0.00000245 = extensión / 12

0,00000245 * 12

0,0000294 m

7 0
2 years ago
Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing
Charra [1.4K]

Answer:

Kinetic energy is given by:

K.E. = 0.5 m v²

Susan has mass, m = 25 kg

Velocity with which Susan moves is, v = 10 m/s

Hannah has mass, m' = 30 kg

Velocity with which Hannah moves is, v' = 8.5 m/s

<u>Kinetic energy of Susan:</u>

0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J

<u>Kinetic energy of Hannah:</u>

0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J

Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.

Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.

4 0
2 years ago
A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and
scoray [572]

Answer:

F_{allow} = 208.15kN

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

K_{allow} =\frac{K_c}{N}

We know that K_c is 33Mpa*m^{0.5} and our Safety factor is 2,

K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

\alpha = 0.5*L_c = 0.5*4mm = 2mm

<em>For the panel</em>

\gamma = 0.5*W = 0.5*250mm = 125mm

To find now the goemetry factor we need to use this equation

\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1

That allow us to determine the allowable nominal stress,

\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}

\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

F_{allow}=\sigma_{allow}*L_c*W

F_{allow} = 208.15*250*4

F_{allow} = 208.15kN

That is the maximum tensile load before a catastrophic failure.

4 0
1 year ago
A circuit is made of 0.4 ohm wire, 150 ohm bulb and a 120ohm rheo stat connected inseries. Determine the total resistance of the
UNO [17]
Answer:
Total resistance of the circuit is 270.4 ohm

Explanation:
We are given that:
resistance of wire = 0.4 ohm
resistance of bulb = 150 ohm
resistance of rheostat = 120 ohm

We are also given that these components are connected in series. This means that the total resistance is summation of all the series components.

Therefore
Total resistance = 0.4 + 150 + 120 = 270.4 ohm

Hope this helps :)
8 0
1 year ago
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?
kaheart [24]

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec )²× 0.20 m

F= 0.034 N

3 0
1 year ago
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