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m_a_m_a [10]
3 years ago
14

A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied

for 2.5s, what is the velocity of the car at the end of the braking period? How far has the car moved during the braking period?
Physics
1 answer:
klemol [59]3 years ago
3 0

Given :

Initial velocity, u = -15 m/s.

Acceleration , a = 2 m/s².

Time taken to applied brake, t = 2.5 s.

To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s

Now, distance covered by car is :

s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m

Hence, this is the required solution.

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The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a
zlopas [31]

Answer:

s = 9.6\times 10^{-12}kg/s

Explanation:

Given:

Solute Diffusion rate  = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel  =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel  =0.10 cm

let the Solute Diffusion rate  of new channel = s

now equating the diffusion rate per unit volume for both the channels

\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}

thus,

s = 9.6\times 10^{-12}kg/s

7 0
3 years ago
Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a
yKpoI14uk [10]

Answer:

Wavelength, \lambda=2.94\ m

Explanation:

It is given that,

Speed of radio waves is v=3\times 10^8\ m/s

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

v=f\lambda

\lambda is wavelength

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m

So, the wavelength of WFNX’s radio waves is 2.94 m.

3 0
3 years ago
As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
Umnica [9.8K]
Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

6 0
3 years ago
How many elections are shared between one nitrogen atom and one carbon atom?
Charra [1.4K]

Answer:

3 electrons from nitrogen and 3 from carbon while carbon already has a lone pair along with a negative charge (called cyanide)

8 0
3 years ago
The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35
azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

g= \frac{GM}{R^2}

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

g_{ss} = \frac{GM}{R+h^2}

g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}

g_{ss} = 8.82m/s^2

Therefore the gravitational acceleration at 350km is 8.82m/s^2

5 0
3 years ago
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