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3 years ago

The two measurements necessary for calculating average speed are

Physics

1 answer:

alisha [4.7K]3 years ago

5 0

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

$Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}$

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

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If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

vovikov84 [41]

Answer: wavelength is $1.30 \times 10^8 nm.$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = \frac{c}{f}$

Plug in the values,

$\lambdam = \frac{3 \times 10^8 m/s}{2.30 GHz\frac{10^9 Hz}{1 GHz}}=0.130 m\frac{10^9 nm}{1 m} = 1.30 \times 10^8 nm.$

5 0

3 years ago

Read 2 more answers

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

2 years ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?

padilas [110]

First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER

5 0

2 years ago

A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th

boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0

2 years ago