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3 years ago

A man stands still on a moving walkway that is going at a speed of 0.2 m/s to

Physics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

0.2 m/s east

Explanation:

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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

Use the drop-down menu to complete the statement. One difference between prokaryotic cells and eukaryotic cells is that a eukary

Ahat [919]

Answer:

Nucleus

Explanation: I did the quiz and got it right. Have an amazing day!

8 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago

Which layer lies between the stratosphere and the

Pepsi [2]

The answer is Mesosphere

8 0

3 years ago

Read 2 more answers

A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the

DiKsa [7]

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

19792.5+11.5m = 32416.5

m = 1097.97 kg

Mass of car = 1098 kg

4 0

3 years ago