All categories

3 years ago

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A:

Physics

1 answer:

gizmo_the_mogwai [7]3 years ago

6 0

Answer: B

Explanation:

Because the ramp is too high, the forces aren't enough to push onto the car stuck down.

You might be interested in

In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predi

Elza [17]

Answer:

Check the explanation

Explanation:

From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .

Kindly check the attached image below for the step by step explanation to the question above.

3 0

3 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

What is time write its SI unit

lisabon 2012 [21]

The SI unit of time is second

4 0

3 years ago

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l

Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

Wₐ = 1500 lb

W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

W = mg

m = W / g

mₐ = Wₐ / g

m_b = W_b / g

mₐ = 1500/32 = 46.875 slug

m_b = 125/32 = 35,156 slug

Let's reduce to the english system

v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

p_f = (mₐ + m_b) v

the moment is preserved

p₀ = p_f

mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

v = $\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}$

we substitute the values

v = $\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6$

v = 0.559 v₀ₐ - 26.40 (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

M = mₐ + m_b

M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

K₀ = ½ M v²

final point. When they stop

K_f = 0

The work is

W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

N-W = 0

N = W

the friction force has the expression

fr = μ N

we substitute

-μ W x = Kf - Ko

-μ W x = 0 - ½ (W / g) v²

v² = 2 μ g x

v = $\sqrt{ 2 \ 0.750 \ 32 \ 17.5}$ Ra (2 0.750 32 17.5

v = 28.98 ft / s

3 0

3 years ago