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vovikov84 [41]
3 years ago
14

A jogger takes five minutes to run a distance of three kilometres. His speed, in metres per second, is approximately A) 5.5 M/S

B) 10 M/S C) 36 M/S D) 4.5 M/S
Physics
1 answer:
just olya [345]3 years ago
5 0

Answer:

His speed, in metres per second, is approximately 10m/s. The correct option is B.

Explanation:

Speed is defined as the rate in which an object cover a distance in a given time. It is measured in meters per second ( m/s). From the question, the jogger covered a distance of 3Km which when converted to meter is 3000meters in a given time 5 minutes which is 300seconds. The calculation of his speed in meter per second is shown below:

Distance= 3Km to meters ( as 1000meters = 1Km). Therefore 3× 1000 = 3000m

Time= 5 minutes to seconds ( as 1 minute = 60 Seconds). Therefore 5× 60= 300.

Speed= distance/ time

= 3000/300

= 10m/s

Therefore his speed, in metres per second, is approximately 10.

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
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Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
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If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2
Elza [17]

Answer:

= 0.5 m/s²

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore;<em> F = ma , where F is the Force, m is the mass and a is the acceleration.</em>

<em>Thus; a = F/m</em>

<em>but; F = 5 N, and m = 10 kg</em>

<em>  a = 5 /10</em>

    <u>= 0.5 m/s²</u>

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3 years ago
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