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jeyben [28]
3 years ago
11

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

of 1.0s the acceleration of the car is
1. 27 m/s2
2. 4.0 m/s2
3. - 4.0 m/s2
4. -12 m/s2
5. -24 m/s2
Physics
1 answer:
wariber [46]3 years ago
4 0

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

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Answer:

1.C

2.C

3.C

Explanation:

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Why might video games increase creativity while the use of cell phones, the internet, or computers do not?
taurus [48]

Explanation:

Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.

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3 years ago
Which of the following will cause an increase in gas pressure in a closed container?
Shalnov [3]
The answer is A. Or the first option. Pressure is changed by lowering the pressure, not reducing the volume. You would assume its C but its A. 
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3 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e
Natasha2012 [34]

Answer:

The total displacement is 102 km 51^o north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

d_n=34km+79km*sin(35^o)=79km

and

d_e=79*cos(35^o)=65km

The total displacement is given by:

D=\sqrt{(79km)^2+(65km)^2}=102km

with an agle of:

\alpha =arctg(\frac{79km}{65km})=51^o

4 0
3 years ago
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