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2 years ago

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On a foggy day, you are driving at 70 mph and Granny is driving at 50 mph. As you try to pass her, you both see an overturned tr

uck and simultaneously slam on your brakes. Assume that both vehicles are identical and that Granny stops just before the truck. Approximately how fast are you traveling when you hit the truck?
a. 35 mph
b. 50 mph
c. 20 mph
d. 0 mph
e. Not enough information

1 answer:

Vitek1552 [10]2 years ago

7 0

Answer:

C?

Explanation:

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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

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

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

Bricks and insulation are used to construct the walls of a house. The

ira [324]

Answer:

$\dot Q=350.438\ W$

Explanation:

Given:

<u>the thermal resistance in the form of </u>

$R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}$

$R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}$

where:

$x_1\ \&\ x_2$ are the thickness of the respective bricks

$k_1\ \&\ k_2$ are the respective coefficient of conductivity

temperature inside the house, $T_h=24\ ^{\circ}C$

temperature outside the house, $\ T_c=10^{\circ}C$

area of the wall, $A=20\ m^2$

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

<u>Using Fourier's law:</u>

$\dot Q=k.A.\frac{dT}{x}$

$\dot Q={dT}\div {\frac{x}{k.A} }$

in series the resistances get add up

$\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )$

$\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )$

$\dot Q=350.438\ W$

7 0

2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0

2 years ago