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3 years ago

What is the effect of changing the net force on the acceleration of an object? Propose a hypothesis.

2 answers:

7 0

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

s344n2d4d5 [400]3 years ago

4 0

Answer:

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

Explanation:

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The current, 2019, women's world record for the

sesenic [268]

Answer: D

8.403 m/s

Explanation:

Given that the women's world record for the 400-meter run around an oval track is held by Marita Koch, with a time of 47.60

seconds.

Her average velocity during this race will be

Velocity = distance/ time

Velocity = 400 / 47.6

Velocity = 8.403 m/s

Therefore, her average velocity during this race will be 8.403 m/s

D is therefore the correct answer.

8 0

3 years ago

Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m

Phantasy [73]

Answer:46.05 mm

Explanation:

Given

$Power=260\ hp\approx 260\times 746=193.96\ KW$

speed $N=550\ rev/min$

allowable shear stress $(\tau )_{max}=15\ kpsi\approx 103.421\ MPa$

Power is given by

$P=\frac{2\pi NT}{60}$

$193.96=\frac{2\pi 550\times T}{60}$

$T=3367.6\ N-m$

From Torsion Formula

$\frac{T}{J}=\frac{\tau }{r}-----1$

where J=Polar section modulus

T=Torque

$\tau$ =shear stress

For square cross section

$r=\frac{a}{2}$

where a=side of square

$J=\frac{a^4}{6}$

Substituting the values in equation 1

$\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}$

$a=0.04605\ m$

$a=46.05\ mm$

7 0

3 years ago

A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent

Margarita [4]

Answer:

$F= 1333.767\,N$

Explanation:

The velocity of the swimmer just before touching the water is:

$v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}$

$v \approx 6.569\,\frac{m}{s}$

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

$(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J$

$F= 1333.767\,N$

6 0

3 years ago

Read 2 more answers

The coefficient of kinetic friction between Gary's shoes and the wet kitchen floor is 0.09. If Gary has a mass of 65 kg, what is

mariarad [96]

Answer:

57 N

Explanation:

Draw a free body diagram. There are three forces:

Weight force mg pulling down.

Normal force N pushing up.

Friction force F pushing horizontally.

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Friction force is the product of normal force and coefficient of friction:

F = Nμ

F = mgμ

F = (65 kg) (9.8 m/s²) (0.09)

F = 57.3 N

Rounded, the friction force is 57 N.

8 0

3 years ago

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul

Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

$\alpha = \frac{\omega_{f}- \omega_{i}}{t}$ Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity (rad/s)

ωi : initial angular velocity (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

$\alpha = \frac{20 -0}{0.65}$

α = 30.77 rad/s²

Newton's second law in rotation:

F = ma has the equivalent for rotation:

τ = I * α Formula (2)

where:

τ : It is the net torque applied to the body. (N*m)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the net torque acting on the CD

Data

α = 30.77 rad/s²

I = 3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

3 0

3 years ago