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gulaghasi [49]
3 years ago
7

If you lift a 3.2kg punkin to a height of 0.80 m, how much work do you do to lift the pumpkin?

Physics
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

Lay the tape over the pumpkin, and to the ground at the blossom end. Try to place it straight down to the ground. Over the Top Side to Side: Place your tape measure on the ground at the side of your pumpkin at it's widest point. Lay the tape over the pumpkin to the other side of the pumpkin, placing the tape on the ground on each side.

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Please help This very important for me you will get 40 points if you help me and explain in deatil please.
posledela
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4 0
3 years ago
Read 2 more answers
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

I_{dB}=10\log(\dfrac{I}{I_{0}})

For one amplifier,

I_{1}=10\log(\dfrac{130}{I_{0}})...(I)

For other amplifier,

I_{2}=10\log(\dfrac{200}{I_{0}})...(II)

For difference in dB levels

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})

I_{2}-I_{1}=10\log(\dfrac{200}{130})

I_{2}-I_{1}=1.870\ dB

Hence, The sound level will be 1.870 dB louder.

7 0
3 years ago
A 29.1 liter tank contains ideal helium gas at 35.8°C and a pressure of 21.8 atm How many moles of gas are in the tank?
crimeas [40]

Answer:

No. of moles, n = 25.022 moles

Given:

Volume of gas in tank, V = 29.1 l

Temperate of gas, T = 35.8^{\circ} = 273 + 35.8 = 308.8 K

Pressure of gas, P = 21.8 atm

Solution:

Making use of the ideal gas equation which given as:

PV = nRT

where

R = Rydberg's constant = 0.0821 L-atm/mol-K

Re-arranging the above formula for 'n' and putting the values in the above formula:

n = \frac{PV}{RT}

n = \frac{21.8\times 29.1}{0.0821\times 308.8}

n = 25.022

3 0
3 years ago
In order for an ion to have a -1 charge, it must have
Jet001 [13]
Gaining electron makes it “negative”
3 0
3 years ago
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