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3 years ago

Can someone help me and put them in order, I numbered them down so it can be easier to say.

Physics

1 answer:

Alexxx [7]3 years ago

5 0

Answer:

the answer to this question is 2,4,3,1

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FREE BRAINLEIAST FOR FIRST GOGOGOGO!!!!heererer

Kamila [148]

Answer:

ME PLS

Explanation:

6 0

2 years ago

Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

6 0

3 years ago

A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m

Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

$K.E+P.E=P.E+K.E$

$\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2$

$\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}$

Put the value into the formula

$\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}$

$\dfrac{v_{1}^2}{2}=18.62$

$v_{1}=\sqrt{2\times18.62}$

$v_{1}=6.10\ m/s$

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0

3 years ago

An electric heater is rated 1000watts. If a current of 5A pass through the heater, find the value of its resistance.

ICE Princess25 [194]

Answer: R=40 Ω

Explanation: $P=I^{2} R$

R=1000/ (25)=40

5 0

3 years ago

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago