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3 years ago

Find the work w1 done on the block by the force of magnitude f1 = 75.0 n as the block moves from xi = -1.00 cm to xf = 3.00 cm .

1 answer:

5 0

The work done is 3.0 Nm. We can us the equation Work = Force * Distance, where Force = 75.0 N, and distance is xf â€“ xi = 3.00 cm - -1.00 cm = 4.00 cm. Convert centimeters to meters by moving the decimal place to the left by two places to get 0.04 m. Plug these values into the Work equation: Work = Force * Distance Work = 75.0 N * 0.04 m Work = 3.0 Nm

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance = 5 m

a) for friction less

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

$v^2 = u^2 - 2gh$

$v = \sqrt{u^2- 2 g h cos 22^0}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

v = 17.58 m/s

b) considering the friction

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

$v^2 = u^2 - 2gh-2\mu_kmgl$

$v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

v = 17.16 m/s

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3 years ago

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Sunny_sXe [5.5K]

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

4 0

3 years ago

Read 2 more answers

What is a substance?

Anna007 [38]

Answer:

A physical matter

Explanation:

A kind of matter with uniform physical properties

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3 years ago

Glass is transparent to visibile light under normal conditions; however, at extremely high intensities, glass will absorb most o

8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

$E=\frac{hc}{\lambda}$

Where $h$ corresponds to Plank constant (6.626070x10^-34Js), $c$ is the speed of light in the vacuum (299792458m/s) and $\lambda$ is the wavelength of the photon(in this case 800nm).

$E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J$

Tranform the units

$1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV$

The band Gap is 4eV, divide the band gap between the energy of the photon:

$\frac{4ev}{1.549802445eV}=2.508974118$

Rounding to the next integrer: 3.

Three photons are the minimum to equal or exceed the band gap.

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3 years ago

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

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3 years ago