Help with 5-10 please

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a

Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

$x_t(t)=d+\frac{1}{2}a_tt^2$

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

$a_t=2.10 m/s^2$ is the acceleration of the truck

The car position instead it is given by the equation

$x_c(t)=\frac{1}{2}a_ct^2$

where

$a_c=3.40 m/s^2$ is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

$x_t(t') = d + 60$

Therefore, solving the equation, we find the time t when this occurs:

$d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s$

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

$x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m$

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

$x_c(t')=x_t(t')$

And by solving the equation, we find the value of d, the initial distance between car and truck:

$\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m$

C)

In order to find the speed of each vehicle, we use the following suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

$a_t = 2.10 m/s^2$

So its speed after t = 7.56 s is

$v_t = 0+(2.10)(7.56)=15.9 m/s$

For the car, we have

u = 0

$a_c=3.40 m/s^2$

So its speed after t = 7.56 s is

$v_c=0+(3.40)(7.56)=25.7 m/s$

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

Which of the following has to happen before the eardrum begins to vibrate with the same frequency as the source of the sound wav

777dan777 [17]

I just took the test it is D

I am 100% sure

8 0

3 years ago

Read 2 more answers

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0

3 years ago

Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc

Sphinxa [80]

Answer:

F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

V_{water} = 1.09 10⁻³ m³

the amount of water is

ρ = m / V

m = ρ V

m = 1000 1.09 10⁻³

m = 1.09 kg

so the weight of the liquid in the container for this time is

W = mg

W = 1.09 9.8

W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v = $\sqrt{0 -2 g ( 0-y_o)}$

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

P₁ = 1.28 10⁵ Pa

The definition of Pressure is

P = F / A

F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

F = P A

F = 1.28 10⁵ 7.46 10⁻⁵

F = 9.5 N

with these data let's use the equilibrium equation

F_ scale -W - F = 0

F_scale = W + F

F_scale = 10.682 + 9.5

F_scale = 20.18 N

4 0

3 years ago

With a mass of 109 kg, Baby Bird is the smallest monoplane ever

OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

m_1=61.4kg
m_2=109kg
v_1=1.84m/s
v_2=?

$\\ \sf\longmapsto ∆P=P$

$\\ \sf\longmapsto m_1v_1=m_2v_2$

$\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}$

$\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}$

$\\ \sf\longmapsto v_2=112.976/109$

$\\ \sf\longmapsto v_2\approx 1.3m/s$

4 0

3 years ago