a car moves at 12m/s and coasts up a hill with a uniform acceleration of -1.6m/s2. how far has it traveled after 6.0s?

Ntally Check for Understanding

Wewaii [24]

Answer: 34

Explanation:

34

3 0

3 years ago

The current in the wires of a circuit is 180.0 milliamps. If the resistance of the circuit were doubled ( with no change in volt

Troyanec [42]

Answer:

I = 0.09[amp] or 90 [milliamps]

Explanation:

To solve this problem we must use ohm's law, which tells us that the voltage is equal to the product of the voltage by the current.

V = I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohm]

Now, we replace the values of the first current into the equation

V = 180*10^-3 * R

V = 0.18*R (1)

Then we have that the resistance is doubled so we have this new equation:

V = I*(2R) (2)

The voltage remains constant therefore 1 and 2 are equals and we can obtain the current value.

V = V

0.18*R = I*2*R

I = 0.09[amp] or 90 [milliamps]

7 0

3 years ago

In a generator, as the magnet spins, opposite poles of the magnet push the electrons in opposite directions. This back-and-forth

Natali [406]

C) alternating current .
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B)direct current </span>

6 0

3 years ago

Read 2 more answers

A car has a force of 0.27 N and an acceleration of 3m/s2 What is the car's mass?

lyudmila [28]

Answer:

0.09 kg

Explanation:

f=0.27

a=3

m=f/a

m=0.27/3

m=0.09

7 0

3 years ago

A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t :

Equations of the uniformly accelerated rectilinear motion of upward (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the initial vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 = (v₀y)²

$v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }$

v₀y = 14.3 m/s

Calculation of the maximum height the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball (R)

We replace data in the equation (1)

x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

$v= \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v= \sqrt{(7.7)^{2}+ (-14.3)^{2} }$

v= 16,2 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })$

$\alpha = tan^{-1} (\frac{-14.3 }{7.7 })$

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0

3 years ago