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3 years ago

A car increases its velocity from zero to 97 km/h in 8.4 s. What is its acceleration?

Physics

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer:

$a=\frac{v-u}{t}$

$v=$ final velocity =97km/h(change to m/s) =26.9m/s NEVER FORGET TO DO THAT BECAUSE TIME IS GIVEN IN SECONDS

$u=$ initial velocity 0 m/s

$t=$ time 8.4s

$a=\frac{26.9-0}{8.4}$

$a=$ $3.2m/2^{2}$

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_____ variables are manipulated by the experiments

nikitadnepr [17]

Answer:

dependent variables

Explanation:

dependent varibeles are the thing you're measuring and independent variables are the thing you change in the exeriment to get a different dependent variable.

may I get brainliest please? :)

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3 years ago

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s

8 0

3 years ago

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e) y = 0 + ½ g t²

t = √ (2y / g)

t = √(2 125 / 9.8)

t = 5.05 s

6 0

3 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

<em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

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3 years ago

In the lesson a thermos is presented as an example of an isolated energy system. How could you change the thermos into an open e

iVinArrow [24]

If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.

7 0

3 years ago

Read 2 more answers