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3 years ago

A scientist wants to use a model to help present results of his detailed scientific investigation

1 answer:

4 0

A problem cause damage damage to people and people who have been doing the damage. A scientist call you when I am done with science

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Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago

What does strong language mean ?

Art [367]

Answer:

noun

the style of a piece of writing or speech.

"he explained the procedure in simple, everyday language"

coarse or offensive language.

noun: strong language

"strong language"

Explanation:

comment how it helps

3 0

3 years ago

Read 2 more answers

A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a

boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, $d_1=3.2\ km$ (due east)

Then the truck moves, $d_2=2.45\ km$ (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

$d=\sqrt{d_1^2+d_2^2}$

$d=\sqrt{3.2^2+2.45^2}$

d = 4.03 km

Let $\theta$ is the direction of the truck’s displacement from the warehouse from south of east.

$\theta=tan^{-1}(\dfrac{2.45}{3.2})$

$\theta=37.43^{\circ}$

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0

3 years ago

A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l

lianna [129]

For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.

3 0

2 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$