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Masja [62]
4 years ago
5

Suppose you increase your walking speed from 5m/s to 15m/s in a period of 1 s. What is your acceleration?

Physics
1 answer:
olchik [2.2K]4 years ago
7 0
Acceleration=change in velocity divided by change in time or a=\frac{\Delta v}{\Delta t}

we are given that it goes from 5m/s to 15m/s
\Delta v=v_f-v_0
in this case, \Delta v=15\frac{m}{s}-5\frac{m}{s}=10\frac{m}{s}

and the change in time or \Delta t is 1 second because this change occurs over a period of 1 second

so the acceleration is a=\frac{10\frac{m}{s}}{1s}=10\frac{m}{s^2}
the acceleration is 10\frac{m}{s^2}
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Acceleration can be defined as the rate at which velocity (blank).
Montano1993 [528]

Rate of change of an object?

6 0
4 years ago
Read 2 more answers
A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?
Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

  • N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω =  11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

  • r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

4 0
3 years ago
Please help
natali 33 [55]

Answer:

Thermosphere

temperature continue to rise up to 1500*C

Ionosphere

Mesosphere

temperature falls to -93*C

Stratosphere

protective ozone layer

UV radiation causes temperature to rise

Troposphere

78% nitrogen, 21% oxygen

temperature drops from 17 to -52*C

Explanation:

3 0
3 years ago
if charge is located at center of spherical volume and electric flux through surface of sphere is Φ what would be flux through s
jonny [76]

Answer:

The flux will be nine times as great.

Explanation:

The electric flux due to a charge Q located in the center of a sphere can be obtained using Gauss's law. Considering a Gaussian surface in the form of a sphere of radius r:

\Phi_E=\int\limits {Ecos\theta dS} \,

The electric field (E) is parallel to the surface vector (dS), so \theta=0

\Phi_E=\int\limits{Ecos(0)dS} \,\\\Phi_E=E\int\limits{dS} \,\\\Phi_E=ES\\\Phi_E=E4\pi r^2

Since the electric flux is proportional to the square of the sphere's radius,  if radius of sphere were tripled, the flux will be nine times as great.

7 0
3 years ago
By what factor does the peak frequency change if the celsius temperature of an object is doubled from 20.0 ∘c to 40.0 ∘c?
mart [117]

Answer:

it increases by a factor 1.07

Explanation:

The peak wavelength of an object is given by Wien's displacement law:

\lambda=\frac{b}{T} (1)

where

b is the Wien's displacement constant

T is the temperature (in Kelvins) of the object

given the relationship between frequency and wavelength of an electromagnetic wave:

f=\frac{c}{\lambda}

where c is the speed of light, we can rewrite (1) as

\frac{c}{f}=\frac{b}{T}\\f=\frac{Tc}{b}

So the peak frequency is directly proportional to the temperature in Kelvin.

In this problem, the temperature of the object changes from

T_1 = 20.0^{\circ}+273=293 K

to

T_2 = 40.0^{\circ}+273 = 313 K

so the peak frequency changes by a factor

\frac{f_2}{f_1} \propto \frac{T_2}{T_1}=\frac{313 K}{293 K}=1.07

8 0
4 years ago
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