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4 years ago

45. An object with a mass of 25 kg accelerates east at a rate of 7 m/s? What fiet force is acting on the object?

Physics

1 answer:

ZanzabumX [31]4 years ago

6 0

Answer:

According to newton's second law of motion F=ma so

Explanation:

Data:-M=25kg ,a=7m/s² F=? ,solution:-F=ma ,F=25×7=175kgm/s² or N

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The voltage ???? in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance ???? is slowly inc

zimovet [89]

Answer:

The change in current at $R =456 \Omega$ is $\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

Explanation:

From the question we are told that

The resistance is $R = 465 \Omega$

The current is $I = 0.09A$

The change in voltage with respect to time is $\frac{dV}{dt} = - 0.03 V/s$

The change in resistance with time is $\frac{dR}{dt} = 0.03 \Omega /s$

According to ohm's law

$V = IR$

differentiating with respect to time using chain rule

$\frac{dV}{dt} = I \frac{dR}{dt} + R * \frac{dI}{dt}$

substituting value at R = 456

$-0.0327 = 0.09 * 0.03 + 456* \frac{dI}{dt}$

$\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

6 0

3 years ago

usted / el partido de fútbol Usted va al partido de fútbol. Question 1 with 1 blankyo / la biblioteca Question 2 with 1 blanknos

34kurt

Answer:

1- Yo VOY A la biblioteca

2- Nosotros VAMOS A la piscina

3- Mis primas VAN AL gimnasio

4- Tú VAS de excursión

5- Ustedes VAN AL parque municipal

6- Alejandro VA AL museo de ciencias

Explanation:

Las preguntas refieren en su totalidad a la conjugación temporal del verbo "ir". Así, refiere a conjugar dicho verbo en tiempo presente simple, el cual se conjuga de la siguiente manera:

Yo voy, el va, tú vas, nosotros vamos, ellos van, ustedes van

Así, debe interpretarse la persona o personas que realizan la acción, para determinar así como se conjuga dicho verbo.

4 0

3 years ago

आप शोर मत करिए। किस प्रकार का वाक्य है- *

Law Incorporation [45]

Answer:

477547748jkdsjdjk

Explanation:

7 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

gavmur [86]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

8 0

3 years ago

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