When drawing a free body diagram, which of the following things would NOT be included

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and

slava [35]

Answer:

f=-1380N

Explanation:

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and as it strikes the board, his hand has a velocity of 23.3 m/s. The master contacts the board for 0.0050 seconds

.the concluding part to the question should be

What is the impact force (impulse) on the board?

solution

from the Newton's second law of motion which states that

the rate of change in momentum is directly proportional to the force applied

f=m(v-u)/t

f=0.3(0-23.3)/0.005

f=-1380N

f=force impact

m=mass of the karates master's hand

t=time for the impact

v=0m/s final velocity

u=initial velocity

6 0

3 years ago

A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc

Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

Given the following data;

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

Angle = 41.7° Northeast.

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

Vr = 7.08m/s

6 0

2 years ago

You and your friend are having a discussion about weight. He claims he weighs less on the 100th floor of a building that he does

balu736 [363]

No the acceleration of gravity on earth is still 9.8ms-2 unless he was on a different planet

7 0

3 years ago

What is the difference between an induced and a permanent magnet?

MissTica

INDUCTION MOTOR:-

Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control

Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation

Power density:-Induction produced by squirrel cage rotor inherently limits power density

Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos

Cost:-Relatively modest initial cost; higher operating costs

PERMANENT MAGNET MORTOR:-

speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range

Reliability:-Lower operating temperatures reduces wear and tear, maintenance • Extends bearing and insulation life • Robust construction for years of trouble-free operation in harsh environments.

power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.

Accuracy:-Without feedback, can be difficult to locate and position to the pinpoint accuracy of servomotors

Cost:-Exhibit higher efficiency, so their energy use is smaller and full return on their initial purchase cost is realized more quickly

8 0

3 years ago

Question Part Points Submissions Used A pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below th

postnew [5]

Answer:

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N

Explanation:

In a problem of impulse and shocks we must use the impulse equation

I = dp = pf-p₀ (1)

p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

V₀ = -19 m / s

θ₀ = 40.0º

Vf = 46.0 m / s

θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

Vcx = Vo cos θ

Voy = Vo sin θ

Vox= -19 cos (40) = -14.6 m/s

Voy = -19 sin (40) = -12.2 m/s

Vfx = 46 cos 30 = 39.8 m/s

Vfy = 46 sin 30 = 23.0 m/s

a) We already have all the data, substitute and calculate the impulse for each axis

Ix = pfx -pfy

Ix = m ( vfx -Vox)

Ix = 0.200 ( 39.8 – (-14.6))

Ix = 10.88 N s

Iy = m (Vfy -Voy)

Iy = 0.200 ( 23.0- (-12.2))

Iy= 7.04 N s

In vector form it remains

I = (10.88 i^ + 7.04 j^) N s

b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

I = Fpro Δt

In the first interval

Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

Fpro = (Fm +0) / 2

In the second interval the force is constant

Fpro = Fm

In the third interval

Fpro = (0 + Fm) / 2

Let's replace and calculate

I = Fpro1 t1 +Fpro2 t2 +Fpro3 t3

I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³

I = Fm 24 10⁻³ N s

Fm = I / 24 10⁻³

Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N

3 0

3 years ago