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2 years ago

A car goes from 0 to 60 mi/h north in 5 seconds. what is the car's acceleration?

Physics

2 answers:

Tomtit [17]2 years ago

8 0

Answer:

<h2>17.6 feet / second ^2</h2>

Explanation:

Explanation: 0 to 60 miles per hour means 0 to 88 feet/sec in five seconds (60 multiplied by 5280 (feet in one mile) divided by 3600 (seconds in an hour). The acceleration is 17.6 feet / second ^2.

OLEGan [10]2 years ago

7 0

Answer:

43200mi/h^2

Explanation:

time will taken in hours so 5/60/60 --> 1/720 h

accelaration formula : (change in velocity or speed) /time

acceleration = (60 - 0)/(1/720) = 43200 mi/h^2

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What is the magnitude of the electric field at the dot in the figure? (figure 1)

blondinia [14]

The magnitude of the electric field at the dot is : 10⁴ v/m

<h3>What is electric field?</h3>

Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

Determine the magnitude of the electric field at the dot

change in voltage = E .d

$100 - 0 = E \times ( 1 \times 10^{-2}\ m ) ----- ( 1 )$

From equation ( 1 )

The magnitude of electric field will be given as

$E = \dfrac{100 }{ ( 1 \times 10^{-2})\ }$

$E = 10^4\ \ \frac {v}{ m}$

Hence we can conclude that The magnitude of the electric field at the dot is : 10⁴ v/m

Learn more about electric field :

brainly.com/question/14372859

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2 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

$\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]$

We know that work is defined as the product of force by distance.

$W=F*d$

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

$f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]$

Now solving the equation for the work.

$d=W/F\\d = 50/19.62\\d = 2.54[m]$

4 0

3 years ago

A 50kg skater at rest on a frictionless rink throws a 2kg ball, giving the ball a velocity of 10m/s. Which statement describes t

Andreas93 [3]

0.4m/s in the opposite direction of the ball.

3 0

3 years ago

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5

svlad2 [7]

The magnetic force experienced by the proton is given by
$F=qvB \sin \theta$
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and $\theta$ the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so $\sin \theta=1$ and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
$F=ma$

So we have
$ma=qvB$
from which we can find the magnitude of the field:
$B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T$

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3 years ago

La _______________ abarca de los dos a los doce años. Se caracteriza por un continuo proceso de adaptación motora, cognoscitiva,

Anna11 [10]

Answer:

please do well to ask questions in English. This will help people provide you answers ASAP. Thank you

5 0

3 years ago