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1 year ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Physics

1 answer:

Rama09 [41]1 year ago

7 0

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

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For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

$85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}$

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity ( $v_t$ ), in order to obtain the angular velocity of the wheel:

$v_t=w*R\\w=\frac{v_t}{R}$

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

$w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}$

And now, recalling that $2\pi$ radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by $2\pi$ :

rotational frequency = $\frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}$

6 0

2 years ago

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

Substituting the values in the above equation

v = 300 x cos 90°

= 300 x 0

= 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

5 0

2 years ago

An object executes simple harmonic motion with an amplitude A. (Use any variable or symbol stated above as necessary.) (a) At wh

valentina_108 [34]

Answer:

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2

7 0

2 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

Flauer [41]

Answer:

The new resistance becomes half of the initial resistance.

Explanation:

The resistance of a wire is given by :

$R=\dfrac{\rho L}{A}$

$\rho$ = resistivity of material

L and A are linear dimension

If the electrical wire is replaced with one having every linear dimension doubled i.e. l' = 2l and r' = 2r

New resistance of wire is given by :

$R'=\dfrac{\rho L'}{A'}$

$R'=\dfrac{\rho (2L)}{\pi (2r)^2}$

$R'=\dfrac{1}{2}\dfrac{\rho L}{A}$

$R'=\dfrac{1}{2}R$

The new resistance becomes half of the initial resistance. Hence, this is the required solution.

4 0

2 years ago

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A littl

ICE Princess25 [194]

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

3 0

2 years ago