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2 years ago

4

A driver traveling at 100\,\dfrac{\text{km}}{\text{h}}100 h km 100, start fraction, start text, k, m, end text, divided by, st

art text, h, end text, end fraction notices the speed limit changes to 50\,\dfrac{\text{km}}{\text{h}}50 h km 50, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction. The driver takes 0.9\,\text{s}0.9s0, point, 9, start text, s, end text to slow down to 50\,\dfrac{\text{km}}{\text{h}}50 h km 50, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction with constant acceleration. We want to know how many meters the driver travels while slowing down.

1 answer:

Xelga [282]2 years ago

3 0

Answer:

67.5

Explanation:

d = (Vo+ v/2)t

d = (150/2)(0.9)

d = 67.5

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A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the

Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

$\Delta P=P_{b}-P_{t}$

Where, $P_{b}$ = Pressure at bottom

$P_{t}$ = Pressure at top

put the value into the formula

$\Delta P=1133-594$

$\Delta P=539\ Pa$

Using formula of pressure for density

$\Delta P = \rho g h$

$\rho =\dfrac{\Delta P}{gh}$

Where, $\rho$ = density

P = pressure

h = height

Put the value in to the formula

$\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}$

$\rho =1100\ kg/m^3$

Hence, The density of the fluid is 1100 kg/m³.

4 0

2 years ago

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0

2 years ago

Read 2 more answers

A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at

Stells [14]

The correct answer is <span>3) $K_f = \frac{1}{2}mv_0^2 + mgh$ .
</span>
In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span> $E=U_i+K_i=mgh + \frac{1}{2}mv_0^2$
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
</span> $E=K_f$ <span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
</span> $K_f = mgh + \frac{1}{2}mv_0^2$ <span>
</span>

7 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco

Vanyuwa [196]

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

$mgh+\dfrac{1}{2}mu^2=mgh'+0$

$100=mgh'$

At lowest point,

$mgh'=K$

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).

7 0

2 years ago