Write this number in standard notation. 4.702 x 10

Natalie lifts a 15-kg rock from the ground onto a 1.5 meter high wall. what is the amount of potential energy she has given the

Zina [86]

The amount of gravitational potential energy acquired by the rock is equal to:
$\Delta U = mg \Delta h$
where
m is the mass of the rock
g is the gravitational acceleration
$\Delta h$ is the increase in height of the rock

Substituting the data of the problem, we find
$\Delta U=(15 kg)(9.81 m/s^2)(1.5 m)=220.7 J$
So, Natalie gave 220.7 J of energy to the rock.

4 0

3 years ago

a carbon atom with 6 proton and 6 neutron in its nucleus is called carbon-12 the carbon atom with 8 neutrons is called what

daser333 [38]

Answer:

carbon-14 must be the answer

4 0

3 years ago

A cannon Is pointed Upward 30 degrees and fires a cannonball with a speed of 100m/sec what is the component of the cannon balls

Kamila [148]

Answer:

86.6 m/s

Explanation:

vₓ = v cos θ

vₓ = 100 m/s cos 30°

vₓ = 86.6 m/s

6 0

3 years ago

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval, the angular velocity d

Nikitich [7]

Answer:

The angular displacement of the wheel is 45 radians

Explanation:

Given;

initial angular velocity, ω₀ = 20 rad/s

final angular velocity, ωf = 10 rad/s

time interval, t = 5

Angular acceleration is calculated as;

$\alpha = \frac{\omega _f - \omega_0}{t} \\\\\alpha = \frac{10 -20}{5} \\\\\alpha = -2 \ rad/s^2$

|α| = 2 rad/s²

Angular displacement is calculated as;

$\theta = \omega_0 \ + \ \frac{1}{2} \alpha t^2\\\\\theta = 20 \ + \ \frac{1}{2} *(2)*5^2\\\\\theta = 20 \ + 25\\\\ \theta = 45 \ radians$

Therefore, the angular displacement of the wheel is 45 radians

8 0

3 years ago

A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for

Elena-2011 [213]

Answer:

1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

In the vertical direction, as the bomb is dropped, its initial velocity is 0.
So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

$\Delta y = \frac{1}{2}*a*t^{2} (1)$

where Δy = -500 m (taking the upward direction as positive).
a=-g=-9.8 m/s²
Replacing these values in (1), and solving for t, we have:

$t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s$

The time required for the bomb to reach the earth is 10.1 s.

2)

In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

$x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.$

3)

The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

$v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s$

5)

If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0

4 years ago

Write this number in standard notation. 4.702 x 10–4