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Thepotemich [5.8K]

2 years ago

7

Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive b

all straight up (from your perspective), how will the speed of the cart change?
a. The speed of the cart will increase
b. The speed of the cart will decrease
c. The speed of the cart will not change
d. You need to know how fast the ball was thrown

1 answer:

Alex2 years ago

6 0

The correct answer is A

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Which statement describes a digital signal used to store information?

velikii [3]

Answer:

D. It is easy to copy.

Explanation:

It is not B, or A, nor C.

8 0

2 years ago

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

$x = 1.17 \ cm = 1.17 \times 10^{-2}\ m$

m = 149 kg

$\delta = 7.87 \ g/cm^3$

$da = 2.28 \times 10^{-10}\ m$

$F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\$

∴

$k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m$

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

$N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2$

$N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2$

$N_{chain} = 2.77 \times 10^{13}$

$N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}$

$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

$\mathbf{k_{si} =45.46 \ N/m}$

4 0

2 years ago

Which of the following statements describe the transfer of energy a. Collision of atoms causing nuclear reactions B. A car speed

oksano4ka [1.4K]

I think the answer would be A.

8 0

3 years ago

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

dolphi86 [110]

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$

5 0

3 years ago

On the graph of voltage versus current, which line represents a 2.0 Ω resistor?

Vikki [24]

Answer:

<h2>line B</h2>

Explanation:

According to ohm's law V = IR where;

V i sthe supply voltage (in volts)

I = supply current (in amperes)

R = resistance (in ohms)

In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

R = 10.0/5.0

R = 2.0ohms

Since the slope of line B is equal to 2 ohms, this shows that the line B is the one that represents the 2ohms resistor.

3 0

3 years ago