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1 year ago

Pls solve this numerical.

Physics

1 answer:

harkovskaia [24]1 year ago

4 0

Answer:

You should write the vectors in their proper sequence: i , j, k

If the vectors are perpendicular their dot product must be zero:

1) A dot B is obviously zero since A and B have no common terms in i , j , k

2) looks like A = 2 i - 4 j + 3 k and B = i + 2 j + 2 k

A dot B = (2 * 1) + (-4 * 2) + (3 * 2) = 0

Note that the dot product produces a scalar with no direction (vector components)

Only i dot i, j dot j, and k dot k have components i dot j = zero, etc.

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A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ( $v_{i}$ ) = 20.0 m/s

Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)

Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0

3 years ago

When you step on the accelerator to increase the speed of your car, the force that accelerates the car is: A. the force of your

Dmitry [639]

Answer:

B. the force of friction of the road on the tires

Explanation:

Unless the car engine is like jet engine, the main force that accelerates the car forward is the force of friction of the road on the tires, which is ultimately driven by the force of engine on the tires shaft. As the engine, and the shaft are part of the system, their interaction is internal. According to Newton laws of motion, the acceleration needs external force, in this case it's the friction of the road on the tires.

6 0

2 years ago

Explain how a persons mother and father can both exhibit or show a trait but their children may not exhibit the trait

beks73 [17]

People have dominant and recessive traits. If there are more dominant traits or the recessive trait just was barely there the parents might not show it while their child could show that recessive trait

4 0

3 years ago

what is the acceleration of each body of mass 5kg rests on a frictionless table and is connected to a cable that passes over a p

Oksanka [162]

Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

T = m₁a (1)

m₂g - T = m₂a (2)

substituting T = m₁a in equation 2:

m₂g - m₁a = m₂a

m₂a + m₁a = m₂g

a(m₁ + m₂) = m₂g

a = m₂g / (m₁ + m₂)

a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²

Both objects have an acceleration of 6.53 m/s²

8 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Pls solve this numerical.​

Pls solve this numerical.