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4 years ago

14

Jen falls out of a tree and lands on a trampoline. The trampoline sags 60 cm before launching Jen back into the air. At the very

bottom, where the sag is the greatest, is Jen’s acceleration upward, downward, or zero?

1 answer:

Ray Of Light [21]4 years ago

7 0

Answer:

At the very bottom, whnere the sag is the greatest, Jay’s acceleration is upward.

Explanation:

As Jay lands on the trampoline, Jay’s motion was dowward, the trampoline was opposing his motion and hence, caused him to reach an initial halt position. Afterwards, the trampoline causes Jay to move back into the air and therefore, the change in velocity vector act in upward direction. The acceleration vector is always align towards the change in velocity vector's direction.

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3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration

Alexus [3.1K]

Answer:

a) a = 7.72 m / s², N = 19.9 N and b) F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

sin52 = Wx / W

cos52 = Wy / W

Wx = W sin52

Wy = w cos 52

Let's write them equations

X axis

Wx = ma

Y Axis

N-Wy = 0

N = Wy

a) Let's calculate the acceleration

a = W sin52 / m = mg sin 52 / m

a = g sin 52

a = 9.8 sin52

a = 7.72 m / s²

The force of the ramp is normal

N = Wy = mg cos 52

N = 3.3 9.8 cos 52

N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

F - Wx = 0

F = Wx

F = mg sin52

F = 3.3 9.8 sin 52

F = 25.5 N

Parallel to the plane and going up

3 0

3 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

4 years ago

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

Marizza181 [45]

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

8 0

3 years ago

(Brainliest) How are radio waves used in cell phone wireless communication technology?

poizon [28]

Radio waves are used in cell phone wireless communication by this: Radio waves are released from the cell phone and travel to satellites in the stratosphere. After they reach the satellite, they are redirected to the recipient of the call/text.

Hope this helps! Can I have brainliest please?

5 0

3 years ago

Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is

goblinko [34]

Answer:

d $\frac{x}{l}$ = m× λ⇒ d = λ ×m×l / x

= 630× $10^{-9}$ m × 3×3m/ 45× $10^{-3}$ m

= 1.26× $10^{-4}$ m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0

3 years ago