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Lerok [7]
3 years ago
9

A child is pushing a shopping cart at a speed of 1.5 m/s how long will it take this child to push the cart down and I'll with th

e length of 9.3 m
Physics
2 answers:
Gelneren [198K]3 years ago
7 0
(9.3 meters) divided by (1.5 meters/sec) = 6.2 seconds .

Here's a suggestion:
When you dictate some text into a machine, take a few seconds
to check and see what the machine thinks you said.
maks197457 [2]3 years ago
4 0
The answer to this question is 6.2 seconds
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Physics questions please help me
Ahat [919]
Hi,Find  answers from Task 5

1.(X+4)+(X)+(X+4)+(X)=50cm

4x+8=50cm

4x=42

X=10.5cm

Length=10.5+4=14.5cm

Width=10.5cm

Area= length × width=(10.5/100) × (14.5/100) =0.0152m2

2. Volume of a sphere= 4/3 ×π×r³

4/3 ×π×r³=3.2×10^-6 m³

r³=3.2×10^-6 m³/1.33×π

r³=7.64134761e-7

r=0.00914m

Surface area of the blood drop= 4πr²

=4×3.142×0.00914×0.00914=0.00105m²

3.

Equation of an ideal gas     = PV =n RT  

Equation for pressure, = P= n RT/V

Equation for the volume of an ideal gas= V= n RT/P

If the volume of gas doubles ,V(new)=  2n RT/P

Equation for temperature of an ideal gas, T = PV/n R

If temperature of gas triples, T (new)= 3PV/n R

New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)

Pressure factor increase= P(new)/P(old)  ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}

=3PV²/2n RT
5 0
3 years ago
Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A
dangina [55]

Answer: action forc roketorce

reaction force is engine fires

4 0
3 years ago
a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The
forsale [732]
The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
6 0
3 years ago
Many television sets show 25 images, called 'frames, each second. What is the time interval between one
Firlakuza [10]

Answer:

Given,

Frame rate = 25 frames per second

To find,

Time interval between one frame and the next.

Solution,

We can simply solve this numerical problem by using the following process.

Now,

Number of frames = 25

Total time taken to display the given number of frames (ie. 25 frames) = 1 second

To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.

So,

Time interval between one frame and next :

= Time taken to display total number of frames / Total frames

= 1/25

= 0.04 second

Hence, time interval between one frame and next is 0.04 second.

8 0
2 years ago
The maximum energy a bone can absorb without breaking is surprisingly small. For a healthy human of mass 60 kg60 kg, experimenta
netineya [11]

Answer:

<em>the maximum height a man can jump from and land rigidly upright on both feet without breaking his legs is 0.34 m</em>

<em></em>

Explanation:

Mass of a healthy man = 60 kg

energy the bone can take without breaking = 200 J

If a healthy man jumps from a height 'h', he falls with an energy equal to the potential energy due to his initial height above the ground.

initial potential energy of the healthy man = mgh

where m = mass of the man

g = acceleration due to gravity = 9.81 m/s^2

h = the height above ground

==> PE = 60 x 9.81 x h = 588.6h

If we assume that all energy is absorbed in the leg bones in a rigid landing, then we can safely say that this calculated PE for a healthy man is equal to the energy his bone can absorb in the jump without breaking.

equating, we have

200 = 588.6h

<em>the maximum height a man can jump from without breaking his legs = 200/588.6 = 0.34 m</em>

When people jump from a height, the sudden deceleration to zero can impact a big force on the leg bones, shattering them. If the time spent in decelerating to zero is increased, the overall force on the leg bones is reduced greatly.

<em>Bending the knees gradually on landing from a jump from a height, and then rolling increases the time spent decelerating, and reduces the impact force on the legs due to the landing</em>. If you observe carefully you will see that this is what professional stunts men and acrobats do when they jump from a height.

5 0
3 years ago
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