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Serjik [45]
3 years ago
11

A small car and an SUV are at a stoplight. The car has amass equal to half that of the SUV, and the SUV's engine canproduce a ma

ximum force equal to twice that of the car. Whenthe light turns green, both drivers floor it at the sametime. Which vehicle pulls ahead of the other vehicle after afew seconds?
a) It is a tie.
b) The SUV
c) The car
Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:a) It is a tie

Explanation:

Given

mass of car is half of SUV

if mass of SUV is 2m so the mass of car is m

SUV can produce twice the force of car i.e. 2 F

so acceleration experienced by SUV

a_{SUV}=\frac{2F}{2m}

a_{SUV}=\frac{F}{m}

acceleration of Car

a_{car}=\frac{F}{m}

both start with zero initial velocity therefore after few second they will be having same velocity and share same position.

                                 

You might be interested in
A 250 g beach ball rolls across the sand with a speed of 11.16 km/h. First convert units to kg and m/ then determine the momentu
yuradex [85]

Answer:

0.775 kg-m/s

Explanation:

Convert the units to the right unit forms necessary

250 g -> 0.25 kg

11.16 km/h -> 3.1 m/s

Now use the formula:

                         velocity

                   mass /  

momentum  /    /

       \           /    /

         \       /    /

          p = mv

p = 0.25 × 3.1 = 0.775 kg-m/s

Hope this helps you!

Bye!

6 0
3 years ago
The rate of change of momentum of a body free falling under gravity is equal to its? A. Velocity B. kinetic energy C. power D. w
jolli1 [7]

Answer:

weight

Explanation:

this is because of the lighter the object the faster it will be affected

7 0
2 years ago
Read 2 more answers
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti
Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}

F_{net}=\sqrt{F^2+4899.78+36.232F}

On differentiating w.r.to F

(\dfrac{dF_{net}}{dF})^2=2F+36.232

0=2F+36.232

F=-\dfrac{36.232}{2}

F=-18.116\ lb

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

6 0
3 years ago
What is weight in Newton’s, of a 50.-kg person on earth
Svetlanka [38]

Answer:

It is about 490 Newtons. 490.3325 to be exact

Explanation:

Pls mark as the Brainliest answer. Thank You very much, enjoy your day

8 0
3 years ago
Highway transportation officials are trying to determine if the average speed on Archer road is above 45mph, the post speed limi
USPshnik [31]

Answer:

Option D) -0.0707

Explanation:

We are given the following in the question:  

Population mean, μ = 45 mp

Sample mean, \bar{x} = 44.9

Sample size, n = 50

Sample standard deviation, s = 10

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{44.9 - 45}{\frac{10}{\sqrt{50}} } = -0.0707

Thus, the test statistic is

Option D) -0.0707

4 0
3 years ago
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