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grigory
20 days 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?
Physics
1 answer:
serg [2.5K]20 days ago
5 0
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
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How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n
inna [2205]

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

7 0
1 month ago
A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph
Softa [2029]
B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.
4 0
3 days ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [2029]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
1 month ago
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [2205]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
25 days ago
Below you are given data about a wave in three different substances.
inna [2205]

1) The wave's period remains constant across different media

2) The wave's velocity varies depending on the medium it travels through

3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged

Clarification:

1)

The period of a wave signifies the duration it takes for one full oscillation.

The wave's period is the inverse of its frequency:

T=\frac{1}{f}

where

T denotes the period

f is the frequency

The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that

f = 350 Hz

thus the period equals

T=\frac{1}{350}=2.86\cdot 10^{-3} s = 2.86 ms

2)

The velocity of a wave can be derived from the wave equation:

v=f \lambda

where

f indicates the frequency

\lambda is the wavelength

<pin the="" first="" medium="">

f=350 Hz, \lambda = 0.75 m, resulting in a speed of

v_1 = (350)(0.75)=262.5 m/s

In the second medium,

f=350 Hz, \lambda = 0.70 m, leading to a speed of

v_2 = (350)(0.70)=245 m/s

In the third medium,

f=350 Hz, \lambda = 0.65 m, showing a speed of

v_3 = (350)(0.65)=227.5 m/s

As a result, we conclude that the wave's speed varies with the medium.

3)

<pwhen a="" wave="" shifts="" from="" one="" medium="" to="" another="" the="" following="" occurs:="">

- The wave's direction alters. Specifically, if the subsequent medium is of greater optical density, the wave bends towards the normal; conversely, it bends away if the second medium is of lesser optical density.

- The wave's speed is affected. The wave decelerates in media with higher optical density and accelerates in those with lower optical density.

- The wave's frequency remains unchanged.

- Ultimately, the wave's wavelength is modified. If moving into a medium of greater optical density, the wavelength decreases, while it increases in one of lower optical density.

Discover more about waves here:

</pwhen></pin>
7 0
16 days ago
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