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Yuliya22 [10]
4 years ago
9

A particle moves so that its position (in meters) as a function of time (in seconds) is r = i ^ + 4t2 j ^ + tk ^. Write expressi

ons for (a) its velocity and (b) its acceleration as functions of time.
Physics
2 answers:
mario62 [17]4 years ago
6 0

Answer:

a.V=8tj+k

b.a=8j

Explanation:

Given:

Position r= i+4t^2j +tk

Nb r is position in metre and time in seconds

a.velocity is change in position/ change in time

v= ∆r/∆t =dr/dt

V=d ( i+ 4t^2j+tk)/dr

Differenting with respect to (t)

V=8tj+K

b.acceleration = change in velocity/change in time

a= ∆v/∆r =dv/dt

a=d (8tj+k)/dt

a= 8j

GREYUIT [131]4 years ago
3 0

Answer:

(a) velocity, v = 8t j + k

(b) acceleration, a = 8 j

Explanation:

The position of the particle as a function of time is given as;

r = i + 4t² j + t k         --------------------(i)

(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;

v = dr / dt = 0 + 8t j + k

v = dr / dt = 8t j + k

v = 8t j + k          ----------------------(ii)

Therefore, the equation/expression for the particle's velocity (v) is

v = 8t j + k

(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;

a = dv / dt = t j + 0

a = dv / dt = t j

a = 8 j

Therefore, the expression for the particle's acceleration, a, is a = 8 j

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Vlad [161]
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t

Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C

Case II
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When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C

Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C

Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C

Case V
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When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
6 0
3 years ago
Air in a thundercloud expands as it rises. If its initial temperature is 300 K and no energy is lost by thermal con- duction on
Ymorist [56]

Answer:

T_{f}=227K

Explanation:

Given data

Initial temperature of air Ti=300 K

Vf (final volume of air)=2Vi (initial volume of air )

To find

Final temperature

Solution

Since no energy is lost by thermal conduction during the process we categorize  this process as an adiabatic one.Hence we apply;

T_{i}V_{i}^{Y-1}=T_{f}V_{f}^{Y-1}\\T_{f}=T_{i}(\frac{V_{i}}{V_{f}} )^{Y-1}\\

Where air is predominantly a diatomic gas Y=1.4m

Substitute the given values

So

T_{f}=300K(\frac{V_{i}}{2V_{i}} )^{1.4-1}\\T_{f}=227K

5 0
4 years ago
What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A
MrRa [10]

Answer:

15.34 kVA

Explanation:

A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.

The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.

Hence Power (P) =√3 × voltage (V) × current (I)

P = √3 × V × I

Given that voltage (V) = 460 V, current (I) = 17 A. Hence:

P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA

But 1000 VA = 1 kVA. Hence:

P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA

8 0
3 years ago
A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the
xxMikexx [17]

Answer:

Time of race  = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

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         17 = 0 x t + 0.5 x 3.89 x t²

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Velocity after 2.96 seconds

          v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

          t=\frac{83}{11.50}=7.22s

Time of race = 2.96 + 7.22 = 10.18 s

3 0
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Heat energy goes from
kifflom [539]

Answer:  Heat moves in three ways: Radiation, conduction, and convection.

8 0
3 years ago
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