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3 years ago

Which object will sink in freshwater, which has a density of 1.0 g/cm3?

Physics

2 answers:

Alecsey [184]3 years ago

7 0

Answer:

The answer is B.

Explanation:

Pani-rosa [81]3 years ago

6 0

Answer:

Object 2, which has a density of 1.9 g/cm3, since it has more density than freshwater.

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The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t

lidiya [134]

Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

Given:

Speed of the car, v = 38.0 m/s.
Speed limit of the highway, $\rm v_o=75.0\ mi/h.$

<h2>Converting the speed limit from mi/h to m/s:</h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

$\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.$

Therefore,

$\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.$

Clearly,

$\rm v_o$

which means, the car driver is exceeding the given speed limit.

6 0

3 years ago

The ___ energy in a mechanical system is determined by adding the potential and kinetic enters together

Svetllana [295]

Answer:

A, total.

The total energy in a mechanical system is determined by adding the potential and kinetic enters together.

i hope this helped at all.

4 0

3 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

3 years ago

If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amper

Elza [17]

Emf = d (phi-B) / dt
B dA/dt, where dA/dt is the area swept out by the wire per unit time. 
0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so 
L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters

6 0

3 years ago

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