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2 days ago

7

Barker is unloading 20kg bottles of water from this delivery truck when one of the bottles tips over and slides down the truck r

amp that is inclined at an angle of 30 degrees to the ground. what amount of force moves the bottle down the ramp?

1 answer:

Maru [2.3K]2 days ago

8 0

<span>The primary force acting on the bottle is gravity. Since the incline is at 30 degrees, the full force of gravity does not apply as the ramp prevents the bottle from descending straight down. By calculating the sine of 30 degrees, it can be determined that the gravitational force acting on the bottle is 4.9 meters per second, and with the 20 kg weight of the bottle, the force acting on it is 98 Newtons.</span>

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A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [2205]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

4 0

16 days ago

A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution

Maru [2355]

Answer:

Explanation:

Amount of gold deposited = 0.5 g

Gold's molar mass = 197 g/mol

Time duration, t = 6 hours

= 6 × 3600

= 12600 s

Calculation of moles: mass/molar mass

= 0.5/197

= 0.00254 mole

Assuming

Au --> Au+ + e-

Faraday's constant = 9.65 x 10^4 C mol-1

Charge, Q = 96500 × 0.00254

= 244.924 C

Relation: Q = I × t

Thus, I = 244.924/12600

= 0.011 A

= 11.34 mA.

6 0

8 days ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

kicyunya [2264]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak $mg-T=\frac{mv^2}{r}$

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as $mg-0=\frac{mv^2}{r}$

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at the bottom $E_b=\frac{mv_0^2}{2}$

By comparing the two states using conservation of energy, we find $v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

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6 0

16 days ago

Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} E ⃗ =(4000 j ^ +3000 k ^ ) N/

kicyunya [2264]

Answer:

Electric flux is calculated as $\phi=31562.63\ Nm^2/C$

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is $E=(4000j+3000k)\ N/C$

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

$\phi=E{\cdot}A$

$\phi=(4000j+3000k){\cdot}Ak$

Applying properties of the dot product, we calculate the electric flux as:

$\phi=3000\times Ak$

$\phi=3000\times \pi (1.83)^2$

$\phi=31562.63\ Nm^2/C$

Consequently, the electric flux for the circular area is $\phi=31562.63\ Nm^2/C$ . Thus, this represents the required answer.

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22 days ago

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Yuliya22 [2420]

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9 days ago