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BabaBlast [244]
3 years ago
10

1. A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly hori

zontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration
Physics
1 answer:
Dovator [93]3 years ago
8 0

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

<em>A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?</em>

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force (F), measured in newtons:

\Sigma F = F - R -f = F_{net} (1)

Where:

R - Resistance force, measured in newtons.

f - Wind force, measured in newtons.

F_{net} - SUV net force, measured in newtons.

If we know that F = 2500\,N, R = 500\,N and f = 500\,N, then net force experimented by the SUV is:

F_{net} = 2500\,N-500\,N-500\,N

F_{net} = 1500\,N

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

a_{net} = \frac{F_{net}}{m} (2)

Where m is the mass of the SUV, measured in kilograms.

If we know that F_{net} = 1500\,N and m = 3500\,kg, then the net acceleration of the car is:

a_{net} = \frac{1500\,N}{3500\,kg}

a_{net} = 0.429\,\frac{m}{s^{2}}

The net acceleration of the SUV is 0.429 meters per square second due west.

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Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal
liraira [26]

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

7 0
3 years ago
A wire has resistivity of 0.12 homes per metre​.
gulaghasi [49]

Answer:

(a) The resistance of 25m of wire is 3 ohms

(b) the length of this wire that has resistance 22 ohms is 183.33 m

Explanation:

Given;

resistivity of the wire, ρ = 0.12 ohms per meter

(a) The resistance of 25m of wire is calculated as follows;

R = \rho L\\\\R = 0.12 \ \frac{ohms}{m} \times 25\ m\\\\R = 3 \ ohms

(b) the length of this wire that has resistance 22 ohms is calculated as;

L = \frac{R}{\rho} \\\\L = \frac{22 \ ohms }{0.12 \ ohms/m} = 183.33 \ m

3 0
3 years ago
8. The fact that voltage can be created by exerting force on a crystal is used in which type of sensor?
AfilCa [17]

Option C

The fact that voltage can be created by exerting force on a crystal is used in Knock sensor

<u>Explanation:</u>

Any knock to an engine exhibits as a little shake that is distinguished by the knock sensor. This sensor acts by altering the fluctuation to an electrical sign, which is later transferred to the processor mastering the ignition system.

There the variation in quake to the voltage sign modifies the timing improvements on the kindling. The knock sensor is placed on the engine base, cylinder cap or consumption manifold. This is because its purpose is to sense fluctuations affected by engine knock or explosion.

5 0
3 years ago
Please help me this is important!
Mariulka [41]
145 Grams!
It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong
4 0
3 years ago
The wing of an airplane experiences the forces as depicted in the vector diagram to the right. Using both one and two dimensiona
Vedmedyk [2.9K]

Answer:

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.

6 0
3 years ago
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