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3 years ago

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A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time int

erval, the car traveled ________ meters.

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

16

Explanation:

$\frac{1}{2} \times 2 \times {4}^{2} = 16$

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Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

KengaRu [80]

Answer:

incomplete question, resistor must be there

Explanation:

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3 years ago

A 6.0-kg ball is sliding to the right at 25.0m/s and strikes a second ball (15-kg) that is initially at rest head-on. After the

katrin [286]

Answer:

15 m/s to the right

Explanation:

Let's say right is positive and left is negative.

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(6.0 kg) (25.0 m/s) + (15 kg) (0 m/s) = (6.0 kg) (-12.5 m/s) + (15 kg) v₂

v₂ = 15 m/s

5 0

3 years ago

6. A 1,780kg car moving at 14m/s takes a turn around a circle with a radius of 50.0m.

m_a_m_a [10]

1) Centripetal acceleration: $3.9 m/s^2$

2) Centripetal force: 6942 N

Explanation:

1)

For an object moving in uniform circular motion (=circular motion with constant speed), the net acceleration is the centripetal acceleration, directed towards the centre of the trajectory and whose magnitude is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the radius of the circle

For the car in this problem, we have

v = 14 m/s is the speed

r = 50.0 m is the radius

Substituting, we find the acceleration:

$a=\frac{14^2}{50.0}=3.9 m/s^2$

2)

The net force acting upon the car is the centripetal force, also acting towards the centre of the circular path, and whose magnitude is given by

$F=ma$

where

m is the mass

a is the centripetal acceleration

For the car in this problem, we have:

m = 1780 kg is the mass

$a=3.9 m/s^2$ is the acceleration

Substituting, we find the centripetal force:

$F=(1780)(3.9)=6942 N$

Learn more about centripetal acceleration and force:

brainly.com/question/2562955

#LearnwithBrainly

6 0

3 years ago

A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill is 8.0 m above the ground. The rocket takes 0.15

sveta [45]

Answer:

$v_i = 18.86 m/s$

Explanation:

As we know that the speed of the rocket is v1 and v2 at the bottom and top of the window

then we will have

$d = (\frac{v_1 + v_2}{2}) t$

$2 = (\frac{v_1 + v_2}{2})(0.15)$

$26.67 m/s = v_1 + v_2$

also we know that

$v_2 - v_1 = (-9.81)(t)$

$v_2 - v_1 = (-9.81)(0.15) = -1.47$

now we have

$v_2 = 12.6$

also we have

$v_1 = 14.1 m/s$

now if the sill of the window is at height 8 m from the ground then we have

$v_1^2 - v_i^2 = 2 a h$

$(14.1^2) - v_i^2 = 2(-9.81)(8)$

$v_i = 18.86 m/s$

8 0

3 years ago

A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force

sasho [114]

Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

5 N = 1 kg * a
a = 5 m/s²

List out known variables:

v₀ = 0 m/s
a = 5 m/s²
v = ?
Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

v² = (0)² + 2(5)(250)
v² = 2500
v = 50 m/s

The final velocity of the box is C) 50 m/s.

7 0

3 years ago