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3 years ago

Lee pushes horizontal with a force of 79 N on a 34Kg mass for 10 m across a floor. Calculate the amount of work lee did

1 answer:

3 0

First you have to know the formula for work which is force times (x) distance which is equal to Newtons times (x) mass answer is in joules which is tge unit

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PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)

melisa1 [442]

D.)downward toward the center of earth

7 0

3 years ago

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

Nataly_w [17]

Answer:

(i) Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

4 0

3 years ago

If an object is accelerating to the right , then net force on the object must be directed towards the right ! True Or False ?

vichka [17]

Not so fast.

I think you're using 'accelerating' to mean 'speeding up', but you really need
to be more careful with it. "Acceleration" means ANY change in speed OR
direction.

If an object's speed to the left is decreasing, or its speed to the right is
increasing, then the net force on the object must be directed towards
the right.

If an object is moving with constant speed in a circular path, then it's
constantly accelerating, because its direction is constantly changing.
The force on it is always directed towards the center of the circle, so
there's one point on the path where the force is directed straight to the right.

8 0

3 years ago

Read 2 more answers

A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.1 cm/min. At what rate is the volume

Ne4ueva [31]

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : $\dfrac{dr}{dt}=-0.1\text{cm/min}$

We know that the volume of a sphere is given by :-

$V=\dfrac{4}{3}\pi r^3$

Differentiating on the both sides w.r.t. t (time) ,w e get

$\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2$

When r= 8 cm

$\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384$

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

6 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago