What's the answer, remember it says no friction

A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent

Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0

3 years ago

How is most of the electricity we use at home generated?

Sveta_85 [38]

Nuclear power plants, wind farms, water farms, and geothermal heating

3 0

3 years ago

John drives a distance of 90 Km to the right at a steady speed of V1=25.00 m/s. Then he stops at the gas station for 10 minutes.

aleksley [76]

a. Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

b. $t=6000\ s$ is the total time taken in the trip

c. $d=126000\ m$ is the total distance

d. $s=54000\ m$ towards right from the starting point.

e. $v_a=21\ m.s^{-1}$

f. $\vec v_a=9\ m.s^{-1}$ towards right.

Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

Total time of round trip when we're given:

distance travelled to the right, $d_r=90000\ m$

speed while travelling to the right, $v_r=25\ m.s^{-1}$

time spent at gas station, $t_g=600\ s$

time spent while travelling back towards the left, $t_l=30\times 60=1800\ s$

speed while travelling to the left, $v_{_l}=20\ m.s^{-1}$

Now time taken for travelling towards right:

$t_r=\frac{d_r}{v_r}$

$t_r=\frac{90000}{25}$

$t_r=3600\ s$

Therefore total time taken in the round trip:

$t=t_r+t_l+t_g$

$t=3600+600+1800$

$t=6000\ s$

c.

Now, distance travelled towards left:

$d_l=v_{_l}\times t_l$

$d_l=20\times1800$

$d_l=36000\ m$

Therefore total distance:

$d=d_l+d_r$

$d=36000+90000$

$d=126000\ m$

d.

Now, total displacement:

$s=d_r-d_l$

$s=90000-36000$

$s=54000\ m$ towards right from the starting point.

e.

Average speed:

$v_a=\frac{d}{t}$

$v_a=\frac{126000}{6000}$

$v_a=21\ m.s^{-1}$

f.

Average velocity:

$\vec v_a=\frac{s}{t}$

$\vec v_a=\frac{54000}{6000}$

$\vec v_a=9\ m.s^{-1}$ towards right.

4 0

3 years ago

Please answer these diagrammatic questions ASAP and please no spam answers

SVEN [57.7K]

Answer:

i. The pressure of due to the water, P, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

3 0

3 years ago

A constant force of 2.0 N is used to push a box 4.0 m across the floor. How much work is done?

nadya68 [22]

Answer:

8 J

Explanation:

Given : Force = 2 N

Distance = 4m

Work done = force * distance

Work done = 2 * 4 = 8 J

6 0

3 years ago