Ask question

Ask question

All categories

4 years ago

13

A hurdler is 0.535 m from a hurdle when he jumps at 6.82 m/s at a 6.79 degree angle. What is his height when he clears the hurdl

e?(Hint: It is not necessarily at the highest point in his motion.) Unit = m. Would love an explanation for this physics problem!

1 answer:

butalik [34]4 years ago

3 0

Answer:

0.82 m

Explanation:

have a good day

You might be interested in

Malia is working with aluminum wire. while working with this type of wire, she must remember to a. always use pressure-type term

Y_Kistochka [10]

Answer:

B

Explanation:

Aluminium doesn't rust unless exposed to copper for a duration of time.

4 0

3 years ago

Often an attacker crafts e-mail attacks containing malware designed to take advantage of the curiosity or even greed of the reci

Mademuasel [1]

Answer:

1) The ethical act Amy should display is to open the Email since it came in through her personal inbox. However, further action should not be taken since ordinarily opening the mail will not disclose or corrupt her system either locally of her system on the server and web space.

2) If such email came in, the first action i will take is to open the mail then take a pause to read through the instruction because most scam or malware comes with an array of further instructions.

After carefully reading through, i will call up Davey to confirm if he sent such mail which answer will obviously be a No and in the case where i am unable to get a call through to Davey, i will immediately delete the mail from my inbox to avoid the mistake of clicking any embedded link within the mail.

5 0

3 years ago

A point in the x-y plane is represented by its x-coordinate and y-coordinate. Design the class Point that can store and process

Black_prince [1.1K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

class pointType

{

public:

pointType()

{

x=0;

y=0;

}

pointType::pointType(double x,double y)

{

this->x = x;

this->y = y;

}

void pointType::setPoint(double x,double y)

{

this->x=x;

this->y=y;

}

void pointType::print()

{

cout<<"("<<x<<","<<y<<")\n";

}

double pointType::getX()

{return x;

}

double pointType::getY()

{return y;

}

private:

double x,y;

};

int main()

{

pointType p2;

double x,y;

cout<<"Enter an x Coordinate for point ";

cin>>x;

cout<<"Enter an y Coordinate for point ";

cin>>y;

p2.setPoint(x,y);

p2.print();

system("pause");

return 0;

}

5 0

3 years ago

Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is

ryzh [129]

Answer:

a) $\dot W = 0.978\,kW$ , b) $I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}$

$COP_{HP} = 14.198$

The reversible work input is:

$\dot W = \frac{\dot Q_{H}}{COP_{HP}}$

$\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$\dot W = 0.978\,kW$

b) The irreversibility is given by the difference between real work and ideal work inputs:

$I = \dot W_{real} - \dot W_{ideal}$

$I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

7 0

4 years ago

The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm

kykrilka [37]

Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

<u><em>1. Force required at the rim of handwheel </em></u>

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw: *d = $\frac{do + dc}{2}$ = (75 + 69) / 2 = 72 mm

and

tan α = p / πd = 6 / (π x 72) = 0.0265

∴ Torque required to overcome friction at he threads is T = P x d/2

T = W tan (α + Ф) d/2

T = $W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}$

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

<u><em>2. Maximum compressive stress in the screw</em></u>

30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

<u><em>3. efficiency of the straightener</em></u>

<u><em></em></u>

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η = To / T = 28,620 / 245,400

η = 0.116 or 11.6%

4 0

3 years ago