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3 years ago

Calculate the acceleration of a bus full speed changes from 6 miles to 12 miles over a period of three seconds

Physics

1 answer:

Jet001 [13]3 years ago

4 0

Answer:

0.90m/s²

Explanation:

Given parameters:

Initial speed = 6miles/hr

Final speed = 12miles/hr

Time taken = 3 seconds

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time. It is mathematically given as:

Acceleration = $\frac{Final speed - Initial speed}{Time}$

We need to convert miles/hr to meters/seconds

Initial speed = 6 x $\frac{miles }{hour}$ x $\frac{1hr}{3600s}$ x $\frac{1609.34m}{1mile}$

= 2.68m/s

Final speed = 12 x $\frac{1609.34}{3600}$ = 5.37m/s

Acceleration = $\frac{5.37 - 2.68}{3}$ = 0.90m/s²

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A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric

lana66690 [7]

The electric flux is 0 N.m²/C.

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

E = 6.20 × 10⁵ N/C

θ = 90⁰

S = 3.20 m²

By substituting the following parameter, we get

Φ = E . Scosθ

Φ = 6.20 × 10⁵ . 3.20cos(90⁰)

Φ = 0 N.m²/C

Hence, the electric flux is 0 N.m²/C

For more on electric flux at: brainly.com/question/26289097

#SPJ4

8 0

1 year ago

Show that the speed of a moving particle over a time interval is constant if and only if its velocity and acceleration vectors a

lawyer [7]

Answer:

$|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C$

$2v(t)\cdot \frac{dv(t)}{dt}=0$

$v(t)\cdot a(t)=0$

Explanation:

Let's start with the definition of a constant velocity.

If the velocity magnitude, in three dimensions, is a constant value (C) we have a constant velocity, which means.

$|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C$

Now, we know that the dot product between v(t) and v(t) is the |v|².

$v(t)\cdot v(t)=|v|^{2}(t)$

If we take the derivative whit respect to time in both sides of this equation we will have:

$\frac{d}{dt}(v(t)\cdot v(t))=\frac{d}{dt}|v|^{2}(t)$

We apply the product rule on the left side and the right side will zero because the derivative of a constant is 0.

$\frac{dv(t)}{dt}\cdot v(t)+v(t)\cdot \frac{dv(t)}{dt}=0$

$2v(t)\cdot \frac{dv(t)}{dt}=0$

We know that dv(t)/dt = a(t) (using the acceleration definiton)

Therefore, we conclude:

$v(t)\cdot \frac{dv(t)}{dt}=0$

$v(t)\cdot a(t)=0$

If the dot product is 0, it means that v(t) and a(t) are orthogonal.

I hope it helps you!

8 0

3 years ago

Does ice melt faster in water or air

STALIN [3.7K]

Answer:

ice melts faster in water

5 0

3 years ago

Read 2 more answers

c) An aluminum tea kettle with mass 1.10 kg containing 1.80 kg of water is placed on a stove. If no heat is lost to the surround

True [87]

Answer:

Q = 640425 J = 640.425 KJ

Explanation:

From the law of conservation of energy:

Heat Addition = Heat Gain of Water + Heat Gain of Aluminum Kettle

$Q = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

Q = Heat addition required = ?

$m_w$ = mass of water =1.8 kg

$m_a$ = mass of aluminum kettle = 1.1 kg

$C_w$ = specific heat capacity of water = = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum kettle = = 890 J/kg.°C

$\Delta T_w$ = $\Delta T_a$ = change in temperature of water and kettle = 95°C - 20°C = 75°C

Therefore,

$Q = (1.8\ kg)(4200\ J/kg.^oC)(75^oC)+(1.1\ kg)(890\ J/kg.^oC)(75^oC)\\$

Q = 567000 J + 73425 J

8 0

3 years ago

Please help !!

nikitadnepr [17]

If the speed is constant, the acceleration a must be zero. Since force F = m•a, the total force must be zero.
In a 1D case, work W is the product of force F and distance d: W = F • d.
Since there is no more information given about friction or air resistance, I have to assume you are looking for the work done by the total force, wich is also zero.

3 0

3 years ago