All categories

3 years ago

A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

Physics

1 answer:

RSB [31]3 years ago

5 0

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

You might be interested in

Help with these questions please , 15 PTS & brainliest

myrzilka [38]

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = $\frac{20}{4}$ = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = $\frac{8}{8}$ = 1A

<h3><u>Alternative method</u></h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

$\frac{R2}{R3}$ = $\frac{I3}{I2}$

$\frac{8}{4}$ = $\frac{I3}{2.5}$

I₃ = 5A

Q6. $\frac{R1}{R2}$ = $\frac{I2}{I1}$

$\frac{4}{8}$ = $\frac{I2}{2}$

I₂ = 1A

6 0

3 years ago

A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?

loris [4]

Answer: -31.36 m/s

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o}+a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

$t=3.2s$ is the time

Knowing this, let's solve (1):

$V_{f}=0+(-9.8m/s^{2})(3.2s)$ (2)

Finally:

$V_{f}=-31.36m/s$ Note the negative sign is because the direction of the bag is downwards as well.

8 0

3 years ago

What is the peacocks displacement from 2 to 3 seconds

mash [69]

1.5 second difference

7 0

3 years ago

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$