Rollita What is the current in the 10.0 , resistor? 10.0 0 120.0 V 320.00 30.0 0

Which of the following statements are true?

inessss [21]

Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

$a = \frac{d\vec v}{dt}$

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

$a = \frac{d\vec v}{dt}$

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

TRUE

Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.

d. If an object's acceleration is zero, its velocity must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

$a = \frac{d\vec v}{dt}$

Since we know that if acceleration is 0 then velocity must be constant

e. If an object's speed is constant, then its velocity must be constant.

FALSE

Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change

7 0

3 years ago

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

0.97c

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = 0.97c

3 0

3 years ago

A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was

Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.