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Mandarinka [93]

3 years ago

15

PLEASE ANSWER...... Suppose a car is traveling in the negative x-direction and comes to a stop. What is the sign of that car’s a

cceleration? How do you know?

2 answers:

Alborosie3 years ago

5 0

positive - accn is slowing car moving in neg x

neg if car in pos x

iren [92.7K]3 years ago

4 0

Answer:

The sign of acceleration is positive

Explanation:

For the car to stop, there must be a deceleration, which will gradually reduce its speed to zero, the deceleration is the aceleration in the opposite way (sign) to the movement, since in this way it will be contrary to the car's displacement and it will be able to stop it.

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Use the drop-down menus to answer each question about hurricanes. What helps create the spin in a hurricane? What is the center

Leto [7]

1.coriolis effect!

2. eye of the storm!

3. warm ocean water!

these should be correct!

6 0

2 years ago

Consider a mass initially moving at 7.50 m/s. How does it take to move 3.5 km (Be sure to convert to meters) if it accelerates a

Tomtit [17]

The mass is moving by uniformly accelerated motion, with initial velocity $v_i=7.50 m/s$ and acceleration $a=0.55 m/s^2$ . Its position at time t is given by the following law:
$x(t)=v_i t + \frac{1}{2}at^2$
where we take the initial position $x_i=0$ since we are only interested in the distance traveled by the mass.

If we put $x(t)=d=3.5 km=3500 m$ into the equation, the corresponding time t is the time it takes for the mass to travel this distance:
$\frac{1}{2}at^2+v_it-d=0$
$4.9t^2+7.5t-3500 =0$
And the two solutions for the equation are:
$t=-25.5 s$ --> negative, we can discard it
$t=24.0 s$ --> this is the solution to our problem

5 0

4 years ago

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Alenkasestr [34]

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

$F=\frac{G\times m_1\times m_2}{r^2}$

G = gravitational constant

$m_1, m_2$ = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

$F\propto \frac{1}{r^2}$

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

8 0

3 years ago

Read 2 more answers

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$

The potential energy reached by the ball at its rebound is

$U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J$

Thus the speed after the collision is

$\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}$

$v'=13.17\ m/s$

The change of momentum is

$\Delta P=0.0616\cdot (13.17+14)$

$\Delta P=1.67 \ kg.m/s$

(b) If the putty sticks to the floor, then v'=0

$\Delta P=0.0616\cdot (0+14)$

$\Delta P=0.86\ m/s$

3 0

4 years ago

At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she

insens350 [35]

Answer:

$1.34 * 10^{3}$ m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/ $m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

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

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$ , we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

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

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$ m/s

8 0

4 years ago