Which is the correct standard notation for 9.3 × 105?

4. An electric iron has a

Licemer1 [7]

Answer:

4.

a) W = 750 J

b) W = 2250 J

c) t = 2 sec

5. Answered in explanation

Explanation:

4.

The formula of power is given as:

P = W/t

where,

P = Power

W = Work Done

t = Time Taken

a)

Here,

P = 750 W

t = 1 sec

W = ?

Therefore,

750 W = W/1 sec

W = 750 J

b)

Here,

P = 750 W

t = 3 sec

W = ?

Therefore,

750 W = W/3 sec

W = (750 W)(3 sec)

W = 2250 J

c)

Here,

P = 750 W

t = ?

W = 1500 J

Therefore,

750 W = 1500 J/t

t = 1500 J/750 W

t = 2 sec

5.

According to Kinetic Particle Theory, the molecules are tightly packed with each other, by strong inter-molecular forces and they can only vibrate at their position. While, molecules or particles in liquids have lesser attractive forces among them. They can move in layers and can take the shape of any container. This is the reason why solid has a definite shape and liquid has none.

6 0

3 years ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

$mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\ v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\ L=0.124m$

6 0

4 years ago

A rocket is launched straight up from the earth's surface at a speed of 1.60×104 m/s . what is its speed when it is very far awa

AlladinOne [14]

16,000 m/s
Since it’s speed, and the distance is unknown. Gravity isn’t applying a noticeable force too on the rocket, as if it were, then the rocket would be accelerating negatively.

7 0

4 years ago

A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for t

Mars2501 [29]

The velocity of the particle is given by the derivative of the position vector:

$\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath$

(a) The particle is moving in the x-direction when the y-component of velocity is zero:

$4ct-2dt = 2t (2c - d) = 0 \implies t=0$

But we want t > 0, so this never happens, unless 2c = d is given, in which case the y-component is always zero.

(b) Similarly, the particle moves in the y-direction when the x-component vanishes:

$2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0$

We drop the zero solution, and we're left with

$c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}$

In the case of 2c = d, this times reduces to t = c/(6c) = 1/6.

7 0

3 years ago

A spring of negligible mass has force constant of 1600 Newtons per meter. (a) How far must the spring be compressed for 3.20 Jou

frez [133]

Answer:

a) x=63.0 $x10^{-3} m$ or 6.3cm

b) x=116.0 $x10^{-3} m$ or 11.6cm

Explanation:

a).

The elastic potential energy is modeling by equation :

$U1=\frac{1}{2}*k*x^{2} \\K=1600 \frac{N}{m}\\ U=3.2J\\m=1.2kg\\x^{2}=\frac{2*U}{k}\\x=\sqrt{\frac{2*3.2J}{1600 \frac{N}{m}}} \\x=\sqrt{4x10^{-3} m^{2}}\\ x=0.06324m$

b).

The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy

$Ug=K2\\mgy=\frac{1}{2}*k*x^{2} \\but \\y=h+x=0.80m+x\\m*g*(0.80m+x)=\frac{1}{2}*k*x^{2}$

Solving for x

$2*(m*g(h+x))=k*x^{2} \\k*x^{2}-(2*m*g*x)-(2*m*g*h)=0\\1600x^{2} -2*1.2kg*9.8\frac{m}{s^{2}}*x-2*1.2kg*0.80m=0\\1600x^{2} -23.52x-18.816m=0$

$x=-\frac{b+/-\sqrt{b^{2}-4*a*c } }{2*a} \\x=-\frac{-23.52+/-\sqrt{-23.52^{2}-4*1600*-18.816 } }{2*a} \\x=11.79 +/- 173.9\\x1=-0.10 m\\x2=0.116$

The negative is discard so

x=0.116m

7 0

4 years ago