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2 years ago

Please help lol im on my sisters acc btw

Physics

1 answer:

Margarita [4]2 years ago

3 0

Answer:

sitting leg curls= hamstring

shrugs=shoulders

machine curls=biceps

bench press= pectorals

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A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

Substitute the given parameters

8.2 = 3.5+1.5a

1.5a = 8.2-3.5

1.5a = 4.7

a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

3 0

3 years ago

Discuss the difference between waveform graphs and vibration graphs

Kitty [74]

Difference exists mainly in the label for x axis.

Explanation:

Shapes of waveform and vibration graphs are same.
Vibration graphs shows the particle at a single location in the path of the wave when time passes.
Waveform graphs shows the particle at multiple locations at a single moment of time.

6 0

3 years ago

Which is belief held by sociologists who work from a social-conflict perspective

Ksju [112]

Some social patterns are helpful, while others are harmful.

7 0

3 years ago

Read 2 more answers

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

3 years ago

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe

Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet = 327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

= 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

= 0.00353 × 128 × (327-40)

= 0.00353 × 128 × 287

= 129.68 J

Quantity of heat required to melt the bullet:

$Q = mH_f$

Q = 0.00353 × 24500

= 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

= 216.17 J

3 0

3 years ago