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2 years ago

A car is traveling at 30m/s it accelerates steadily for 5 s after which it is travelling at 50 m/s calculate its acceleration?

2 answers:

7 0

Formula for acceleration: a = v-v0/t
a is acceleration, v is final velocity and v0 is starting velocity, t is time
Now plug it in
a = 50m/s - 30 m/s = 20 m/s/ 5s = 4 m/s^2
The acceleration is 4 m/s^3

Mashcka [7]2 years ago

4 0

For this question , we should apply
a = v - u by t
Given - u = 30m/s
v = 50m/s
t = 5 secs
Solution :
a = v - u by t
a = 50 - 30 by 5
a = 20 by 5
( cut 5 and 20 because 5 x 4 = 20 )
a = 4 m/s^2

.:. The acceleration is 4 m/s^2

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A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₁² v₂

$v_1= \dfrac{r_1^2}{r_2^2} v_2$

$v_1= \dfrac{1.4^2}{2.8^2} v_2$

$v_1= 0.25 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)$

$\Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)$

$v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}$

$v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}$

v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)² x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0

3 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

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2 years ago

A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

Hopefully this will help you.

4 0

3 years ago

A 1 pound bucket rests on a table. What is the support force exerted on the bucket by the table? What is the support force when

katrin2010 [14]

This question involves the concepts of equilibrium and Newton's third law of motion.

The support force will be "1 pound" for the empty bucket and the support force will be "6 pounds" after pouring water into it.

According to the condition of equilibrium, the sum of forces acting on a stationary object must be zero. Hence, the support force of the table will be equal to the total mass of the bucket.
According to Newton's Third Law of Motion every action force has an equal but opposite reaction force. Hence, the support force will be a reaction force to the weight of the bucket.

Therefore, the support force in each case will be equal to the total mass of the bucket:

Case 1 (empty bucket):

support force = 1 pound

Case 1 (water poured):

support force = 1 pound + 5 pound

support force = 6 pound

Learn more about equilibrium here:

brainly.com/question/9076091

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3 years ago

Un ciclista en una cuesta hacia abajo, marcha a una velocidad de 45km/h aplica los frenos y al cabo de 5s su velocidad se ha red

nexus9112 [7]

Answer:

Los 0.0416km

esto se debe a que transponemos la fórmula acelerada y obtenemos Distancia = velocidad × tiempo

también recuerda transponer los segundos a horas viendo que la velocidad es por hora

También tenga en cuenta que no hablo español, así que esto fue extremadamente difícil

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2 years ago