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nekit [7.7K]
3 years ago
13

A 75.0 kg man sits on a massless cart that is on a horizontal surface. The cart is initially stationary and it can move without

friction or air resistance. The man throws a 5.00 kg stone in the positive direction, applying a force to it so that it has acceleration +3.50 m/s2 as measured by a nearby observer on the ground. What is the man's acceleration during the throw, as seen by the same observer? Be careful to use correct signs.
Physics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:0.233 m/s^2

Explanation:

Given

Mass of man m=75 kg

mass of stone m_0=5 kg

acceleration of stone a=3.5 m/s^2

To provide a acceleration of 3.5 m/s^2[/tex] he must exert a force of

F=m_0\times a=5\times 3.5=17.5 N

same Force is exerted by stone on cart

17.5=m\times a

a=\frac{17.5}{75}=0.233 m/s^2

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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
a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}

Now, we have that

{v_f}^2-{v_0}^2=2a\Delta x

so we end up with a distance traveled of

\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=79.6\,\mathrm m

6 0
3 years ago
A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti
melisa1 [442]
1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

4 0
3 years ago
What is an example of kinetic energy
Olenka [21]

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

6 0
3 years ago
What is the minimum v (in km/s) the rocket engines must provide to allow the craft to escape from the Earth?
coldgirl [10]

Answer:

11 kilometers (7 miles) per second, or over 40,000 kilometers per hour (25,000 miles per hour)

Explanation:

3 0
4 years ago
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