All categories

3 years ago

A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?

1 answer:

8 0

We could determine the acceleration using this formula
$\boxed{a= \dfrac{v_{1}-v_{0}}{t} }$

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
$a= \dfrac{v_{1}-v_{0}}{t}$
$a= \dfrac{0-23}{5}$
$a= \dfrac{-23}{5}$
a = -4.6
The acceleration is -4.6 m/s²

You might be interested in

A 6.2 kg cannonball is fires from a 280 kg cannon. The cannon recoils with a velocity of 4.9 m/s. What is the velocity of the ca

lara [203]

Answer:

666 bc its on its way to my house

Explanation:

3 0

3 years ago

Why are common measurement systems important

kenny6666 [7]

So that international people don't mistake units for others. ex 1 inch could be mistaken for 1 cm in non American countries

6 0

2 years ago

An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

2 years ago

What is the weight of an object (mass = 60 kilograms) on Mars, where the acceleration due to gravity is 3.75 meters/second2?

enot [183]

Answer:

Weight of an object = 225 newtons

Explanation:

Given:

Mass = 60 kilograms

Acceleration due to gravity = 3.75 m/s²

Find:

Weight of an object

Computation:

Weight of an object = Mass x Acceleration due to gravity

Weight of an object = 60 x 3.75

Weight of an object = 225 newtons

5 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 m/s protons (mass m) at a gas target of an unkno

matrenka [14]

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$ .

(a)

Total kinetic energy conservation

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$

Replacing the given data

$m_{u}= \frac{m_{p}(3.6x10^{7}-3.3x10^{7})^{2}}{((3.6x10^{7})^{2}-(3.3x10^{7})^{2})}$

Then,

$m_{u}=0.04245m_{p}$

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

$v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}\\\\v_{u_f}=\frac{m_{p}(3.6x10^{7}-3.3x10^{7})}{0.04245m_{p}}\\\\v_{u_f}=7.067x10^{7} m/s$

5 0

3 years ago