All categories

4 years ago

A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?

Physics

1 answer:

dybincka [34]4 years ago

3 0

Answer:

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle $a=10m/sec^2$

We have to find the radius of the circle

We know that acceleration is given by $a=\frac{v^2}{r}$

So $10=\frac{5^2}{r}$

$r=\frac{25}{10}=2.5m$

So radius of the circle will be 2.5 m

You might be interested in

Transforme 0,4m em mm

rewona [7]

1m = 1000 mm

0,4m = 0,4 * 1000 = 400mm

3 0

3 years ago

An emergency first aid kit should include

AlekseyPX

I say B

hope this helps

6 0

4 years ago

Inessa05 [86]

Savings account is where she can have access to her money anytime in case of an emergency.

<h3>What is a Savings account?</h3>

This is a type of account run by a financial institution which yields interests on deposits made.

This type of account usually comes with features in which the money can be withdrawn via an ATM card or by visiting the bank to collect it. This therefore makes it the most appropriate choice for Analisa.

Read more about Savings account here brainly.com/question/25787382

7 0

2 years ago

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra

Black_prince [1.1K]

Answer:

$\frac{F_1}{F_2}=3.55$

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

$v_2=\frac{2}{5}v_1$

$F_1=\frac{1}{2}\rho_1 CAv_1^2$

$F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2$

Dividing the above two equations we get

$\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55$

The ratio of the drag forces is $\mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}$

5 0

3 years ago

The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.

Natali5045456 [20]

<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION : lnPso2cl12=-kt+lnPso2cl1 initial partial pressure Pso2cl12 the rate constant k and the time t lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375) so lnPso2cl12=3.002 we take the base e antilog: lnPso2cl12=e3.002 Pso2cl12=20 torr we use the integrated first order rate lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002 we use the same rate constant and initial pressure k=4.5*10-2*s-1 Pso2cl12=375 Pso2cl12=1* so2cl12 Pso2cl12=37.5 torr subtract in Pso2cl12 grom both side lnPso2cl12- lnPso2cl12=-kt ln(x)-ln(y)=ln (x/y) ln (Pso2cl12/Pso2cl20)=-kt we get t -1/k*ln(Pso2cl12/Pso2cl20)=t t=51 s</span>

6 0

3 years ago