A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

Em₀ = U = mg y

Final. Low point just before the crash

Emf = K = ½ m v²

Em₀ = Emf

m g y = ½ m v²

v = √ 2 g y

Let's calculate

v = √ (2 9.8 0.05)

v = 0.99 m / s

1) the moment before the crash is

p₀ = m v

p₀ = 0.221 0.99

p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

p = 0

2) change of momentum

Δp = p - p₀

Δp = 0- 0.219

Δp = -0.219 kg m / s

3) the reason is

Δp / p = 1

In percentage form it is 100%

3 0

3 years ago

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ( $\Phi$ ) = 20°

Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

4 0

3 years ago

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A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of

aev [14]

Answer:

initial magnetic field 1.306 T

Explanation:

We have given area of the conducting loop $A=0.065m^2$

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by $e=\frac{d\Phi }{dt}=-A\frac{db}{dt}$

$1.2=0.065\times \frac{db}{0.087}$

$db=1.606T$

So initial magnetic field = 1.606-0.3= 1.306 T

5 0

3 years ago

Introduction to Forces<br> Warm-Up Active<br> How do forces affect the motion of an object?

In-s [12.5K]

Answer:

Forces can affect an object.

Balanced forces allow an object to continue moving at a constant motion (law of inertia).

Unbalanced forces cause a change in motion.

4 0

2 years ago

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The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b

Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, $\mu_o=4\pi\times 10^{-7}\ N.s^2/C^2$

Let $\epsilon_o$ is the permittivity of free space. The relation between $\mu_o,\epsilon_o\ and\ c$ is given by :

$c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}$

$\epsilon_o=\dfrac{1}{c^2u_o}$

$\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}$

$\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2$

Hence, this is the required solution.

3 0

3 years ago

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A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°​

A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°