All categories

3 years ago

What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds

Physics

1 answer:

SIZIF [17.4K]3 years ago

4 0

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

You might be interested in

Power=potantial difference times what

Andru [333]

Answer:

Power=Potential difference times current

Explanation:

7 0

4 years ago

A 10 kilogram sled is pulled across a frictionless surface with a force of 50 newtons for a distance of 10 meters. The pull is a

Gala2k [10]

Answer:

The power will be "250 watt". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 50 N

Displacement,

d = 20 m

Time,

t = 2.0 seconds

Whenever the block is pulled, the angle will be "0" i.e., Cos0° = 1

Now,

The work will be:

= $Force\times displacement\times \Theta$

On substituting the given values, we get

= $50\times 10\times Cos0^{\circ}$

= $50\times 10\times 1$

= $500 \ Newton$

Now,

The Power will be:

= $\frac{Work \ done}{time}$

= $\frac{500}{2.0}$

= $250 \ watt$

8 0

3 years ago

How many elemens are listed in the periodic table

Vesnalui [34]

118 elements
from what knowledge i have gathered in my human brain i hope that this is helpful

5 0

3 years ago

Read 2 more answers

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

$qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s$

So, the speed of the electron is 2591.83 m/s.

8 0

3 years ago

The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago