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3 years ago

6

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number. J

2 answers:

morpeh [17]3 years ago

6 0

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

$PE=mgh$

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

$PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

disa [49]3 years ago

4 0

Answer:

1225J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

1225J

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The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa

My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

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Using formula of Cpk

$Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})$

Put the value into the formula

$Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})$

$Cpk=min(0.83,2.5)$

$Cpk=0.83$

Hence, The value of Cpk is 0.83.

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A 1250 kg car, driving 7.39 m/s, runs into the back of a stationary 5380 kg truck. After the collision, the truck moves forward

Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3$

$v_1=-2.5\ m/s$

So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

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VashaNatasha [74]

Answer:

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Explanation:

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$r_1 = 11.64 \hat i + 43.46\hat j$

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

$r_2 = 30 cos15\hat i + 30 sin15 \hat j$

$r_2 = 28.98\hat i + 7.76 \hat j$

now the displacement of the boat is given as

$d = r_2 - r_1$

$d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)$

$d = 17.34 \hat i - 35.7 \hat j$

so the magnitude is given as

$d = \sqrt{17.34^2 + 35.7^2}$

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Alex73 [517]

Explanation:

Below is an attachment containing the solution.

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