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1 year ago

8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?

1 answer:

pychu [463]1 year ago

5 0

(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²

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A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the

zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂= L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0

2 years ago

Read 2 more answers

Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mu

MArishka [77]

Answer:

Horizontal component: $F_x = 58\ N$

Vertical component: $F_y = 33.5\ N$

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

$F_x = F * cos(angle)$

$F_x = 67 * cos(30)$

$F_x = 58\ N$

For the vertical component, we have:

$F_y = F * sin(angle)$

$F_y = 67 * sin(30)$

$F_y = 33.5\ N$

So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.

4 0

1 year ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

1 year ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

1 year ago

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

1 year ago