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Cerrena [4.2K]
2 years ago
15

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

d your answer to two significant figures
Physics
2 answers:
xxMikexx [17]2 years ago
8 0

Answer: 9130 joules

Explanation:

Workdone by wheelbarrow = ?

Time = 11 seconds

Power = 830 watts

Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

i.e Power = (workdone/time)

830 watts = Workdone / 11 seconds

Workdone = 830 watts x 11 seconds

Workdone = 9130 joules

Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

WITCHER [35]2 years ago
4 0

Answer:

9100

Explanation:

got right on egd

You might be interested in
Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm
ozzi

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

3 0
2 years ago
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an
leonid [27]

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

4 0
2 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
2 years ago
A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the hor
Volgvan

Answer:

h=23.67 m  : Building height

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 12.2 m/s  , at an angle  α=53° above the horizontal

x= 25 m , y=0

Calculation of the time it takes for the ball to hit the ground

We replace data in the equation (1)

x = xi + vx*t

x= 25 m   ,xi=0 ,  vx= v₀*cosα = (12.2 m/s)*cos(53°) =7.34 m/s

25 = 0 + 7.34*t

t= 25 / 7.34

t= 3.406 s

Calculation of the Building height

v₀y =  v₀*sinα = (12.2 m/s)*sin(53°) = 9.74 m/s

in t= 3.406 s, y=0

We replace data in the equation (2)

y= y₀ + (v₀y)*t - (1/2)*gt²

0=  y₀ + (9.74)*(3.406 )- (1/2)*(9.8)(3.406 )²

0=  y₀ + 33.17- -56.84

0=  y₀ - 23.67

y₀ =  23.67 m =h: Building height

5 0
2 years ago
A 500 μF capacitor is wired in series with a 5 V battery and a 20 kΩ resistor. What is the voltage across the capacitor after 20
liq [111]

Answer:

Explanation:

It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.

Q = CV ( 1 - e^{-\lambda\times t} )

λ = 1/CR , C is capacitance and R is resistance.

= 1/(500 x 10⁻⁶ x 20 x 10³ )

= .1

λ t = .1 x 20

λ t = 2

CV = 500 X 10⁻⁶ X 5

= 2500 X 10⁻⁶ C

Q = 2500 x 10⁻⁶ ( 1 - e^{-2} )

= 2500 x 10⁻⁶ x .86566

= 2161.66 μ C .

voltage = Charge / capacitor

2161.66 μ C  / 500μ F

= 4.32 V

3 0
2 years ago
Read 2 more answers
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