All categories

3 years ago

Stan's car starts from rest and accelerates to a speed of 25 m/sec in 10 seconds. What is the acceleration? *

Physics

1 answer:

marta [7]3 years ago

5 0

Answer:

2.5m/s(squared)

Explanation:

u=0,v=25,t=10,a=?

a=change in velocity/time

a=v-u/t

a=25-0=25/10

a=2.5m/s(squared

You might be interested in

What is pulling force expansion

Aloiza [94]

Answer:

physical cosmology, the Big Rip is a hypothetical cosmological model concerning the ultimate fate of the universe, in which the matter of the universe, from stars and galaxies to atoms and subatomic particles, and even spacetime itself, is progressively torn apart by the expansion of the universe at a certain time

3 0

3 years ago

Read 2 more answers

What defines minerals will mark brainlest if croccted right

zvonat [6]

<span> Uniform chemical composition
Solid
Crystalline structure </span>these are the characteristics of minerals

3 0

3 years ago

The chart shows data for a moving object. A 2-column table with 3 rows. Column 1 is labeled time in seconds with entries 2, 4, 6

Rashid [163]

The object is not accelerating. I think I got this question right

3 0

3 years ago

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: $Vf^{2} = Vo^{2} - 2gh$ and $h=Vo*t - \frac{g*t^{2} }{2}$ , First we need to find the height to time iqual to 3.04 s using the formula: $h=Vo*t - \frac{g*t^{2} }{2}$ and remember that golf ball was released from the rest (Vo= 0 (m/s)) so $h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}$ , we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using $Vf^{2} = Vo^{2} - 2gh$ solving for Vf, we get: $Vf = \sqrt{Vo^{2}-2*g*h }$ replacing the values given $Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }$ , so we get: Vf = 29.79 (m/s).

5 0

3 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$