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Zepler [3.9K]
3 years ago
15

A car starting from rest is accelerated uniformly to a velocity of 100.8km/h. Convert this to metre per secondphy​

Physics
2 answers:
Harrizon [31]3 years ago
7 0

Explanation:

100.8 km/h × (1000 m/km) × (1 h / 3600 s) = 28 m/s

Olenka [21]3 years ago
5 0

Answer:

28 m/s

Explanation:

100.8km/hr * 1hr/3600sec* 1000m/km= 100.8*1000/3600 m/s

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An equation for the motion of an object is given as 2vt-4x=Bt^2 where B is a constant. The variable v indicates velocity, in met
Jlenok [28]

Answer:

acceleration m/s²

Explanation:

The units on the left side of the equation are units of length. The units on the right side of the equation must be units of length also. Acceleration times time squared gives length.

8 0
3 years ago
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7.1 Project Guidelines 2021
Ede4ka [16]
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
3 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
4 years ago
An astronaut and his space suit have a combined mass of 157 kg. The
alexgriva [62]

Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]

6 0
3 years ago
What are the three important safety factors for surgical technologists to consider when exposed to ionizing radiation?
Gemiola [76]

Answer:

Exposure time limitation, shielding and distance.

Explanation:

  • Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
  • Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
  • Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.
7 0
3 years ago
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