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4 years ago

Uniform solid disk rolls without slipping down a 19.0° inclined plane. what is the acceleration of the disk's center of mass?

1 answer:

3 0

Below is the solution:

Let us say that the disk goes through a vertical elevation change of one meter.

The change in potential energy will equal the change in kinetic energy 

PE = KEt + KEr 
mgh = ½mv² + ½Iω² 

for a uniform disk, the moment of inertia is 
I = ½mr² 
and 
ω = v/r 

mgh = ½mv² + ½(½mr²)(v/r)² 
mgh = ½mv² + ¼mv² 
gh = ¾v² 

v² = 4gh/3 

v² = u² + 2as 

if we assume initial velocity is zero 

v² = 2as 
a = v² / 2s 

s(sinθ) = h 
s = h/sinθ 

a = 4gh/3 / 2(h/sinθ) 
a = ⅔gsinθ 

a = ⅔(9.8)sin25 
a = 2.8 m/s²

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A cannon tilted upward at 30° fires a cannonball with a speed of 100 m/s. At that instant, what is the component of the cannonba

Sonja [21]

Answer:

the cannonball’s velocity parallel to the ground is 86.6m/S

Explanation:

Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.

For this case we must draw the velocity triangle and find the component in X(see atached image).

V= Initial velocity=100M/S

$cos30=\frac{Vx}{V}$

V= Initial velocity=100M/S

Vx=cannonball’s velocity parallel to the ground

Solving for Vx

Vx=Vcos30

Vx=(100m/S)(cos30)=86.6m/s

the cannonball’s velocity parallel to the ground is 86.6m/S

6 0

3 years ago

The temperature of a heat engine is 500 K. Some of the heat generated by the engine flows to the surroundings, which are

Naddik [55]

Answer:

The efficiency of the engine is 75% under given conditions.

Explanation:

Efficiency is defined as the ratio of work done by the engine to the total heat supplied to a heat engine. We know work done by the heat engine be difference between total heat supplied $\left(Q_{h}\right)$ by total heat utilized $\left(Q_{c}\right)$ . We can derive formula for efficiency by considering the above factors, $\text { efficiency }=\frac{\text { work done }}{\text { heat suplied }}=\frac{Q_{h}-Q_{c}}{Q_{h}}=1-\frac{Q_{c}}{Q_{h}}=1-\frac{T_{1}}{T_{2}} \therefore \text { efficiency }=1-\frac{T_{1}}{T_{2}}$

Now consider a heat engine is at a temperature of 500K and the heat produced by the engine released to the surroundings at a temperature of 125K. The efficiency is calculated by using the above formula, $\text { efficiency }=1-\frac{T_{1}}{T_{2}}=1-\frac{125}{500}=1-0.25=0.75$

∴ Efficiency of the engine is 75%

8 0

3 years ago

John runs for 10 mins at a uniform speed 9kn/h. At what speed should he run for the next 20mins so that theaverage speed comes t

kolbaska11 [484]

Answer:

13.5 km/h

Explanation:

John runs for 10 min at 9 km/h, then 20 min at v. He will have run a total of 30 min at 12 km/h.

Therefore:

(30 min) (12 km/h) = (10 min) (9 km/h) + (20 min) v

v = 13.5 km/h

4 0

3 years ago

A 90-meter train travels at a constant speed of 10 meters per second. How long will it take to cross a 0.06 bridge?

salantis [7]

Answer:

The time taken for the train to cross the bridge is 9.01 s

Explanation:

Given;

length of the train, L₁ = 90 m

length of the bridge, L₂ = 0.06 m

speed of the train, v = 10 m/s

Total distance to be traveled, = L₁ + L₂

= 90 m + 0.06 m

= 90.06 m

Time of motion = Distance / speed

Time of motion = 90.06 / 10

Time of motion = 9.006 s ≅ 9.01 s

Therefore, the time taken for the train to cross the bridge is 9.01 s

5 0

3 years ago

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

4 0

3 years ago