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yanalaym [24]
3 years ago
10

If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

owing equation. S(t) = 76 + 128t − 16t2 (a) What is the average velocity (in ft/sec) in the first 4 seconds after it is thrown? ft/sec (b) What is the average velocity (in ft/sec) in the next 4 seconds?
Physics
1 answer:
faltersainse [42]3 years ago
3 0
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

    s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    Displacement in 4 seconds = 332 - 76 = 256 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

    Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

    Average velocity in second 4 seconds is 63.5 ft/s downward

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________
siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

To know more about escape velocity follow

brainly.com/question/14042253

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8 0
1 year ago
Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is
Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

r = \frac{mv}{Bq}

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}

r = 0.072 m

r = 7.2 cm

5 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both
Nuetrik [128]

Answer:

Part a)

Width of the slit is

a = 580 nm

Part b)

Ratio of intensity is given as

\frac{I}{I_o} = 0.81

Explanation:

Part a)

As we know by the formula of diffraction we will have

a sin\theta = \lambda

so we have

\theta = 90

\lambda = 580 nm

so we will have

a sin90 = 580 nm

a = 580 nm

Part b)

As we know that the intensity in diffraction pattern is given as

I = I_o (\frac{sin\theta}{\theta})^2

\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2

so for angle 45 degree

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = 0.81

7 0
3 years ago
If a 340 N girl sits on a seesaw and is lifted 1.4 m in 1.6 s, how much work was done by the child on the other side?
blondinia [14]
In physics, work is defined as the total energy when an object is moved to a certain displacement by the application of external force. It is calculated by the expression W = Fd. For this case, the displacement is apparently zero, then there is no work in the system above.
4 0
3 years ago
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