Which planet is 92,897,000 miles from the sun?

A hockey player makes a slap shot exerting a constant force of 40.0 newtons on the puck for .2 seconds. What is the magnitude of

Taya2010 [7]

Answer:

Magnitude of impulsive force is 8 N.s.

Explanation:

Given:

Force acting on the puck is, $F=40.0\ N$

Time interval for which the force acts is, $\Delta t=0.2\ s$

We are asked to find the impulsive force.

Impulsive force acting on a body is the sudden change in the momentum of the body due to the application of a large constant force for a very small interval of time.

Here, the hockey player applies a large force of 40 N for a very short interval of time. So, the impact on the puck is measured as Impulse and is represented by 'J'.

Impulse in terms of applied force and time interval is given as:

$J=F\Delta t$

Plug in the given values and solve for 'J'. This gives,

$J=40\times 0.2\\J=8\ N\cdot s$

Therefore, the magnitude of impulsive force is 8 N.s.

6 0

3 years ago

Whats the origin of all stars? a) supernova b) dwarfs c) protostars d) nebulae

Fudgin [204]

All stars originated from c., protostars. Protostars are the earliest phase of the process of star evolution.

-Starry Sky

Oh lmao, my name and the irony. cx

5 0

3 years ago

Read 2 more answers

Please help quick please

Ede4ka [16]

I can’t see the picture what do you need help with

3 0

2 years ago

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

Genrish500 [490]

Answer:

Explanation:

Given

mass of block $m=5.7\ kg$

at $t=0 s$

displacement is $x=-0.7\ m$

velocity $v=-0.8\ m/s$

acceleration $a=2.7\ m/s^2$

suppose $x=A\cos (\omega t+\phi )$ is the general equation of SHM

where A=amplitude

$\omega$ =natural frequency of oscillation

therefore velocity and acceleration is given by

$v=-A\omega \sin (\omega t+\phi )$

$a=A\omega ^2\cos (\omega t+\phi )$

for t=0

$-0.7=A\cos (\phi )---1$

$v=-0.8=-A\omega \sin(\phi)---2$

$a=2.7=-A\omega ^2\cos(\phi )----3$

divide 1 and 3 we get

$\omega ^2=\frac{27}{7}$

$\omega =\sqrt{\frac{27}{7}}$

Now square and 1 and 2 we get

$(0.7)^2+(\frac{0.8}{\omega })^2=A^2$

$A^2=0.49+0.166$

$A=0.81\ m$

3 0

3 years ago

In a domestic electric circuit (220.V),an electric kettle of 3kW power rating is operatedthat has a current rating of 4A. What r

kozerog [31]

Answer:

(i) It will take longer (75/22 times) the time to boil a given quantity of water compared to when the current is working at its rated capacity

(ii) The reasons are;

(1) For economy; most of the appliances in the home only require low power circuitry with thinner wire while a separate high power circuitry is created directly from the main supply for the high electric power rated appliances

(2) For safety; to prevent the over heating of the electric circuits when an high electric power appliance needs to be connected an high power electric power outlet has to be specified

Explanation:

(i) The power rating of the kettle = 3 kW

The voltage rating of the circuit, V = 220 V

The current rating, I = 4 A

The formula for electric power = I² × R = I × V

Therefore, we have;

Power produced = 220 V × 4 A = 880 V·A = 800 W

Hence, since the power produced is below the power rating of the electric kettle, it will take a longer time to boil a given amount of water than specified by the kettle manufacturer

The energy supplied H = V×I×t

Where:

t = Time in seconds

Therefore, we have;

3 kW = 3000 W;

3000 × t₁ = 880 × t₂

t₂/t₁ =3000/880 = 75/22

Hence the kettle will take 75/22 multiplied by the time it takes when working at rating capacity to boil a given quantity of water

(ii) This is so because the power consumption already factored in the electrical installation as well as the type of appliances utilized in the home allow for several low power rating consumption and few high power rating consumption

Therefore, for both economy and safety the electrical circuit are split to allow for the use of very thick copper or aluminium electric cables in the high power rating electric circuits to which can be plugged high electric power consuming devices such as the water heater and electric cooker

The low electric power consuming devices, such as the electric bulb and fans are connected to the low or "regular" power rating electric circuit outlets

The current required for high power and low power appliances is different and also the fuse rating required for both the appliances different, so two separate circuits are used for high power and low power appliances.

6 0

3 years ago