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3 years ago

11

Your car gets a flat! You go from 90 kilometers per hour to a stop in 6 seconds. What is your rate of deceleration? (it's negati

ve!)
I need this ASAP!

1 answer:

defon3 years ago

8 0

Answer:

-15?

Explanation:

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Can someone help me, it will get 5 stars

Tcecarenko [31]

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6 0

3 years ago

Read 2 more answers

Read the scenario. A car travels 25 m/s forward for 10 s. Which option accurately identifies the measurements within the scenari

Phantasy [73]

Explanation:

It is given that,

A car travels 25 m/s forward for 10 s.

Solution,

For a vector, a quantity must have both magnitude as well as the direction. For a scalar, a quantity have only the magnitude. In this case, the car moves in forward direction. This is the only difference between the vector and the scalar.

Out of given option,s the correct option is (c) "The measurement 25 m/s is the only vector quantity because it is a measurement of speed".

5 0

3 years ago

A person just supports a mass of 20kg suspended from a rope.

Georgia [21]

Answer:

F = 200 N

Explanation:

Given that,

The mass suspended from the rope, m = 20 kg

We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,

F = mg

Where

g is acceleration due to gravity

Put all the values,

F = 20 kg × 10 m/s²

F = 200 N

So, the resultant force on the mass is 200 N.

7 0

3 years ago

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

Daniel [21]

The answer is 1575, I just took the Review.<span />

7 0

3 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago