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hammer [34]
3 years ago
5

6.) What force would evetually cause the ball to come to a stop after leaving the ramp​

Physics
1 answer:
solniwko [45]3 years ago
7 0

\colorbox{aqua} A\colorbox{aqua} N\colorbox{aqua} S\colorbox{aqua} W\colorbox{aqua} E\colorbox{aqua} R\colorbox{aqua} S

Forces of inertia

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An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (cons
Nata [24]

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

8 0
3 years ago
1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later
LekaFEV [45]

Answer:

262.5 Nm

Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

\tau = \dfrac{\Delta L}{t}

Δ<em>L</em> is the change in angular momentum.

Using values in the question,

\tau = \dfrac{8800-6700 \text{ kg m}^2\text{/s}}{8\text{ s}}  = 262.5 \text{ Nm}

4 0
3 years ago
A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.
lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, V_{p} = 120\ V          (rms voltage)

Voltage at secondary, V_{s} = 13000\ V  (rms voltage)

Current in the secondary, I_{s} = 8.50\ mA  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}

where

N_{p} = No. of turns in primary

N_{s} = No. of turns in secondary

\frac{N_{s}}{N_{p}} = \frac{13000}{120} ≈ 108

(b) The power supplied to the line is given by:

Power, P = V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW

(c) The current rating that the fuse should have is given by:

\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}

\frac{13000}{120} = \frac{I_{p}}{8.50}

I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A

 

6 0
3 years ago
The brakes on your automobile are capable of creating a deceleration of 4.6 m/s^2. If you are going 114 km/h and suddenly see a
drek231 [11]

Answer:

The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.

Explanation:

a=\frac{V_f-V_0}{t}

isolating "t" from this equation:

t=\frac{V_f-V_0}{a}

Where:

a=-4.6m/s^2 (negative because is decelerating)

V_f= 79 km/h

V_0= 114 km/h

First we must convert velocity from km/h to m/s to be consistent with units.

79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s

114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s

So;

t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s

7 0
3 years ago
Which device or set of devices is contained in a mobile telephone?
Masja [62]

 

Un teléfono móvil o teléfono celular es un dispositivo portátil que puede hacer o recibir llamadas a través de una portadora de radiofrecuencia, mientras el usuario se está moviendo dentro de un área de servicio telefónco.​ El enlace de radiofrecuencia establece una conexión con los sistemas de conmutación de un operador de telefonía móvil, que proporciona acceso a la red telefónica pública conmutada (PSTN). La mayoría de los servicios de telefonía móvil modernos utilizan una arquitectura de red celular, y por lo tanto los teléfonos móviles son, con frecuencia, llamados celulares, especialmente en Hispanoamérica. En España, se utiliza más el término móvil.

6 0
3 years ago
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