Find the momentum of a 3.0 kg mass when it is stopped

A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

Leno4ka [110]

<h3><u>Question</u><u>:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

<h3><u>Statement:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

<h3><u>Solution</u><u>:</u></h3>

Initial velocity (u) = 70 m/s
Acceleration (a) = -14 m/s^2
Time (t) = 3 s
Let the velocity of the car after 3 s be v m/s
By using the formula,

v = u + at, we have

$v = 70 + ( - 14)(3) \\ = > v = 70 - 42 \\ = > v = 28$

So, the velocity of the car after 3 s is 28 m/s.

<h3><u>Answer:</u></h3>

The car's speed after 3 s is 28 m/s.

Hope it helps

3 0

3 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Norma-Jean [14]

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

$v^2 = 2gh = 2*9.8*0.587 = 11.5$

$v = \sqrt{11.5} = 3.4 m/s$

b) Vertical and horizontal components of the velocity are

$v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s$

$v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s$

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

$\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s$

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

$s_h = v_ht = 1.8*0.293 = 0.528 m$

3 0

3 years ago

A 50 mm diameter thin walled pipe is covered with an insulation layer with thicknessof 25mm and thermal conductivity of0.075W/mK

Wittaler [7]

Answer:

The steam will start to condense at 6.6 mm into the pipe

Explanation:

The volume flow rate =π×(50/1000)²/4×10 = 0.0196 m³/s

The specific volume of the steam = 1.769 m³/kg

Therefore;

The mass flow rate = 0.0196/1.769 = 0.011099 kg/s

The resistance of the insulation material = ln(0.075/0.05)/(2×π×0.075) = 0.860 K/W

The resistance of the outside film of the insulator = 1/(15×2×π×0.075×1) = 0.14147 K/W

The total resistance = 0.14147 + 0.860 = 1.00147 K/W

1/(UA) = 1.00147 K/W

A = 2×π×0.05×1

1/U = 0.3146

U = 3.178 W/m² K

We have;

T(x) = T₀ + (Tin - T₀) exp(-UπDx/mcp)

Therefore, when T(x) = 100°C, we have;

100 = 20 + (120 - 20)exp(-3.178×π×0.05x/(0.011099 × 1.33))

Solving, we get

x = 6.597× 10⁻³ m ≈ 6.6 mm

Therefore, the steam will start to condense at 10 mm into the pipe.

3 0

4 years ago

Two small objects are suspended from threads. When the objects are moved close together, they attract one another. What of the f

creativ13 [48]

Answer:d

Explanation:

All the given situations are possible because

(a)When particles are oppositely charged then they attract each other

(b)One is Positively charged and other is uncharged: Charged particle will induce charges of opposite nature to attract the other particle

(c)Negatively charged particles will induce the positive charge in the uncharged particle to attract the initially uncharged particle.

4 0

3 years ago

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

11111nata11111 [884]

Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

we substitute

E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

k = 1 / 4πε₀

E = 2k $\frac{\lambda}{ r}$

5 0

3 years ago