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2 years ago

Which of the following are NOT two of the best ways to StaySafe around electricity?

Physics

1 answer:

creativ13 [48]2 years ago

3 0

1) Sleeping with an electronic in charge right next to you.

2) Leaving a charger plugged in all the time especially near liquid substances.

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Three oxygen isotopes

igomit [66]

Stable isotopes, radioisotopes, oxygen

6 0

2 years ago

A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

8 0

3 years ago

You can tell from reading this passage that -

Romashka-Z-Leto [24]

Answer: B or "diamonds are rare,valuable gems".

Explanation:

since the link does not show very much, I don't know if A is correct or not. D is incorrect because that is an opinion. C is incorrect.

4 0

3 years ago