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Yuri [45]
2 years ago
12

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di

agram shows the forces acting on the cart.
A free body diagram with 3 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up, labeled F Subscript N Baseline. The third vector is pointing up to the right at an angle of 35 degrees, labeled F Subscript p Baseline = 250 N. The up and down vectors are the same length.

The mass of the cart, to the nearest whole number, is
Physics
2 answers:
faust18 [17]2 years ago
6 0

Answer:

146kg

Explanation:

right on edge

Pachacha [2.7K]2 years ago
4 0

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

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A 240 V circuit has 20 A flowing through it. How much power is it using?
Delvig [45]

Explanation:

P = IV

V = 240

I = 20

P = ?

P = 240 × 20

P = 4800 watts

6 0
3 years ago
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
single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

4 0
3 years ago
For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming
Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

7 0
2 years ago
To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?
UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally  placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.



4 0
2 years ago
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