A car traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the cars acceleration as it slows down?

1. A racing car with the driver weighs 1825 lb. Find the kinetic energy in ft*lb when traveling with a speed of 100 mi/hr.

Svetlanka [38]

Answer:

1. 610,000 lb ft

2. 490 J

Explanation:

1. First, convert mi/hr to ft/s:

100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s

Now find the kinetic energy:

KE = ½ mv²

KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²

KE = 610,000 lb ft

2. KE = ½ mv²

KE = ½ (5 kg) (14 m/s)²

KE = 490 J

6 0

3 years ago

The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr

DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

$I = \frac{V}{Z}$

Where;

V is the voltage in the AC circuit

Z is the impedance

$Z = \sqrt{R^2 + X_L^2}$

Where;

R is the resistance

$X_L$ is the inductive reactance

$X_L$ = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

$I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }$

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago

Suppose a greenhouse is to be constructed to keep plants warm in the winter. Which material should be used for the windows of th

Anastasy [175]

Answer:

the answers, material D meets the requested characteristics

Explanation:

The objective of an insulating material for the house, must allow solar radiation to enter, so that the plants can perform photosynthesis and must prevent radiation from inside the house from being lost.

Therefore the material must meet two conditions be transparent to sunlight and be absorbed from the radiation coming from the house; this is to leave for visible light and absorb infrared radiation

Reviewing the answers, material D meets the requested characteristics

4 0

2 years ago

A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is

Maksim231197 [3]

Correct question:

A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.05 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A.

For what values of x will constructive interference occur at point P?

Answer:

values of x in which constructive interference will occur at point P are; 0.775 m, 2.025 m, 3.275 m, 4.525 m, 5.775 m, 7.025 m, 8.275 m

Explanation:

Given;

frequency, F = 120 MH = 120 x 10⁶ Hz

distance between A and B = 9.05 m

make a sketch of this antenna

A-------------------P------------------B

x 9.05 - x

Now, we calculate the wavefront between the two antenna

λ = v/f

where;

v is speed of light = 3 x 10⁸ m/s

λ = (3 x 10⁸) / (120 x 10⁶)

λ = 2.5 m

Thus, the constructive interference = n(2.5)

For constructive interference to occur at point P, then

the path difference = constructive interference

Path difference = 9.05 - x - x

= 9.05 -2x

∴ 9.05 - 2x = n(2.5)

2x = 9.05 - n(2.5)

x = 4.525 - n(1.25)

Finally, determine the value of x for which n = -1, -2, -3 and 0, 1, 2, 3

when n = -1

x = 4.525 + 1(1.25)

x = 5.775m

when n = -2

x = 4.525 + 2(1.25)

x = 7.025m

when n = -3

x = 4.525 + 3(1.25)

x = 8.275m

when n = 0

x = 4.525 - 0(1.25)

x = 4.525m

when n = 1

x = 4.525 - 1(1.25)

x = 3.275m

when n = 2

x = 4.525 - 2(1.25)

x = 2.025m

when n = 3

x = 4.525 - 3(1.25)

x = 0.775m

Thus, values of x in which constructive interference will occur at point P are; 0.775 m, 2.025 m, 3.275 m, 4.525 m, 5.775 m, 7.025 m, 8.275 m

4 0

2 years ago