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4 years ago

Hey can anyone help me out in dis pls!

Physics

2 answers:

mart [117]4 years ago

7 0

Answer:it’s a

Explanation:

patriot [66]4 years ago

3 0

Answer:

Explanation:

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A car of mass 875 kg is traveling 30.0 m/s when the driver applies the brakes, which lock the wheels. The car skids for 5.60 s i

AleksAgata [21]

Explanation:

It is given that,

Mass of the car, m = 875 kg

Initial speed of the car, u = 30 m/s

Brakes are applied i.e. v = 0

The car skids for 5.60 s in the positive x - direction before coming to rest, t = 5.6

(a) Acceleration of the car, $a=\dfrac{v-u}{t}$

$a=\dfrac{0-30}{5.6}$

$a=-5.35\ m/s^2$

(b) Force, F = ma

$F=875\ kg\times -5.35\ m/s^2$

F = -4681.25 N

So, the force of 4681.25 N is acting on the car.

(c) Let x is the distance covered by the car. So,

$v^2-u^2=2ax$

$0-u^2=2ax$

$x=\dfrac{-u^2}{2a}$

$x=\dfrac{-(30\ m/s)^2}{2\times -5.35\ m/s^2}$

x = 84.11 meters

So, the distance covered by the car is 84.11 meters. Hence, this is the required solution.

7 0

3 years ago

A physics teacher performed a demonstration for a science class by pulling a crate across the floor

nika2105 [10]

Answer:

An opposing force caused by friction produced a lower acceleration than calculated.

Explanation:

3 0

3 years ago

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some

erica [24]

If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.

Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

$P=\rho g h$

Here $\rho$ is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$

Learn more about gauge pressure here,

brainly.com/question/14012416

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2 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$