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Nadusha1986 [10]
3 years ago
13

What does inertia mean

Physics
2 answers:
daser333 [38]3 years ago
6 0

Answer:

a property of matter by which it continues in its existing state of rest of uniform motion in a straight line, usless that state is changed by an external force.

abruzzese [7]3 years ago
5 0

a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
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From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle
GarryVolchara [31]

Answer:

Part a)

y = 88.5 m

Part b)

v_x = 7.7 m/s

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

y = v_y t + \frac{1}{2}gt^2

Part d)

x = 30.8 m

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

y = v_y t + \frac{1}{2}a_y t^2

y = 2.5(4) + \frac{1}{2}(9.81)(4^2)

y = 88.5 m

Part b)

components of the velocity is given as

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

y = v_y t + \frac{1}{2}gt^2

Part d)

distance where it will strike the floor is given as

x = v_x t

x = 7.7 \times 4

x = 30.8 m

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

y = v_y t + \frac{1}{2}gt^2

10 = 2.5t + \frac{1}{2}(9.81) t^2

t = 1.2 s

7 0
3 years ago
A car turns into a driveway that slopes upward at a 9 degree angle, car is moving at 6.5 m/s. If the driver lets the car coast,
expeople1 [14]

To put it in the simplest form, the automatic transmission has a torque converter that uses the transmission fluid to turn the other side of the torque converter. Ex. Putting two room fans facing each other and turning one on and wind flowing from one fan makes the second fan facing the rotating fan turn and the idling engine probably doesn't have enough power to hold it.

I guess more technically, in an automatic, the car selects a neutral gear (no gear) when it is at rest, and the brake pedal is pressed. Upon releasing the brake, the car will apply a small forward force that will hold the car steady on a very small gradient or propel it forwards slowly on a flat or downhill gradient. This force is only started when the brake is being released, as this is the indicator for the car to change into 1st gear.

In a situation where a car is rolling back down the hill it is facing up, there could be one or more of several situations at hand.

Most likely, it simply lacks power to hold on the gradient of the slope. In this case, you'll have to perform a hill start to maintain brake force until forward propulsion is enough to move the car forward. Hill starts are almost always necessary for manual transmission cars. Additionally, it could be the driver's pedal transfer from brake to accelerator is to slow. That would also account for some part of the slip.

Okay, what's missing here is the clutch. The clutch as I'm sure you're aware is a series of plates that connect to each other and transfer power due to friction. The amount of friction is adjustable depending on how much force is applied to hold the plates together (or apart). It is easier to see this in a manual car at low speeds, where the driver hovers the clutch actuation pedal around the "Friction Point". This point is the fine line between moving (increased friction between clutch plates holds them together more firmly, thus transferring more power) and staying stationary (clutch plates disengaged from each other). From the Friction Point, any further release of the clutch will cause the car to move forwards because the transmission is engaging with more of the engine's power. Depressing the clutch pedal back in will not have any effect, as it will just keep the clutch plates separate.

In an automatic car, this is all controlled by computer algorithms, determining how much the clutch should be engaged to reach a certain speed. Taking off from the lights on a hill for example will not necessarily register as any different to taking off on flat ground. The effect of this is that the car is assuming that is requires a certain number of revs and a predetermined clutch setting to accelerate smoothly. Due to the increased force the hill provides, the car will move backwards until the power again reaches a level that will overcome its slippage.

One other thing, is that clutches work both ways. Since only friction holds them together, the torque exerted by the wheels back through the drive-train to the clutch can cause the plates to slip when they are not completely engaged. This results in the wheels moving independently of the running engine, as the clutch is separating the forces they would exert on each other.

I think the key thing to note is that despite not having a pedal to operate it, Automatic cars still have a clutch - just one that relies on a computer to function.

If it's not the engine or the clutch which are both behaving as normal, and we're still assuming it's an auto, there could be a problem with the car's computer system, transmission, clutch or gearbox that is causing the slippage. I would assume this is less likely, but it might be worth checking if it happens to your car. your frickin' welcome

6 0
3 years ago
Read 2 more answers
Can you travel faster by not running forward ?
GrogVix [38]
Yeah i think with a car or a plane:)
5 0
3 years ago
Calculate the change in temperature of 20 kg of water if 30 kg of aluminum is dropped in the water and the aluminum changes temp
vivado [14]
 <span>ΔE = 0 

(m_alum) (c_alum) Δt - (m_water)(c_water) Δt' = 0 

(30) (897) (20) - (20)(4186) Δt' = 0 

Δt' = 6.42857 °C</span>
6 0
4 years ago
An image of a car with height of 14 cm occurred in the mirror which is located at a T-
Thepotemich [5.8K]

Answer:

The distance of car form the mirror is 330 cm.

Explanation:

height of object, h = 140 cm

height of image, h' = 14 cm

radius of curvature, R = 60 cm

focal length, f = R/2 = + 30 cm

Let the distance of image is v and the distance of object is u.

\frac{h'}{h}\frac{v}{u}\\\\\frac{14}{140} =\frac{v}{u}\\\\v =\frac{u}{10}

Use the formula of focal length

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{30}=\frac{10}{u}+\frac{1}{u}\\\\\frac{1}{30}=\frac{11}{u}\\\\u = 330 cm

3 0
3 years ago
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