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frez [133]
2 years ago
7

3. A 1500 kg car moving at 30 m/s strikes a 6000 kg van initially at rest. If the car

Physics
1 answer:
MissTica2 years ago
3 0

Answer:

7.5m/s

Explanation:

Force= mass × velocity

Energy is conserved, the car and van should have the same overall force.

1500kg × 30m/s= 6000kg × final velocity

Final velocity = 7.5m/s

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A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli
Vaselesa [24]

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

4 0
3 years ago
4. Three methods that people use are:
Mila [183]
The answers to question 4 d
6 0
2 years ago
What is the mass of a car that has a kinetic energy if 4,320,000 J moving at 23 m/s?
Alekssandra [29.7K]

Answer:

kinetic energy=1/2mv^2.

which is 4320000=1/2×m×23^2.

which is 4320000=1/2×m×529.

4320000=264.5m.

m=4320000/264.5.

m=16332.70~16333g

4 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
A player kicks a football from ground level with an initial velocity of 27.0 30.0 above the horizontal Find the ball's maximum h
stepladder [879]

Answer:

Given

initial velocity (u) =27.030

Force of gravidity g) =9.8

Rtc maximum height Hmix =?

sln \\ \  hmix = u {}^{2}  \div 2g \\ 27.030 \div 2 \times 9 \\  \\ hmix = 38.025m

5 0
2 years ago
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