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balandron [24]
2 years ago
9

100 point question!! A driver begins to brake when her car is traveling at 15.0 m/s, and the car comes to a stop 4.0 s later. Ho

w much farther does the car go after the driver begins to brake, assuming constant acceleration?
Physics
1 answer:
Anni [7]2 years ago
3 0

Answer:

30 m

Explanation:

Let's list out the known variables that we are given in this question.

  • v₀ = 15 m/s
  • v = 0 m/s
  • t = 4 s

We are trying to solve for displacement in the x-direction.

  • Δx = ?

Find which constant acceleration equation contains all 4 of these variables.

  • \displaystyle \triangle x = \Big (\frac{v+v_0}{2} \Big ) t

Plug the known values into the equation and solve for delta x.

  • \displaystyle \triangle x = \Big (\frac{0+15}{2} \Big ) \cdot 4
  • \displaystyle \triangle x = \Big (\frac{15}{2} \Big ) \cdot 4  
  • \displaystyle \triangle x =\frac{60}{2}
  • \triangle x = 30

The car goes 30 m after the driver begins to brake.

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Answer:

80×5×10=4000J

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8 0
3 years ago
Find the force necessary to pull a 6 kg object 3 m/s2
mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

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<h3>18 N</h3>

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6 0
2 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
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Zolol [24]

Answer:

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What must you know to find the amount of work done on an object
Cerrena [4.2K]

Answer:

The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules.

Explanation:

5 0
3 years ago
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