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denis-greek [22]

2 years ago

9

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

how the following will be different for car 2: a) braking distance, b) thinking distance and c) stopping distance.

1 answer:

Zepler [3.9K]2 years ago

8 0

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

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A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m abov

m_a_m_a [10]

Answer:

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²

3 0

3 years ago

is plugged into the outlet of a 120−V circuit that has a 20−A circuit breaker. You plug an electric hair dryer into the same out

WINSTONCH [101]

Answer:

Given the point of maximun electric power of the hair dryer 1500 (w); The circuit breaker won´t trip at all

Explanation:

The most simplyfied relation between power, voltage and current is:

Electric power (in watts) =Voltage (in volts) * current (in ampers)

In the case of P= 1500 (w) and V= 120 (v) we have:

I = 1500/120 (a) = 12,65 (a)

This value is far away of 20 (a) (the nominal trip current

8 0

3 years ago

A Cathode of initial mass 10.00g weigh to 10.05g

Leto [7]

Answer:

i'm sorry i'm not a physics student

6 0

2 years ago

A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$
where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.

6 0

2 years ago

Read 2 more answers

N76 [4]

Answer:

Weight

Explanation:

Weight is the downward pull on an object due to gravity.

For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.

4 0

3 years ago