PHYSICS HIGH SCHOOL

Alejandra weighs 225 newtons. how much work does she do against gravity when she climbs to a ledge at the top of a 15 meter climbing wall

Answers

Answer 1

Answer: Work=Fd
=Fgd
= 225N * 15M
= 3375 Nm

Answer 2

Answer:

Final answer:

Alejandra does 3375 Joules of work against gravity when she climbs to the ledge at the top of the climbing wall.

Explanation:

In order to calculate the work done against gravity, we can use the formula: Work = Force x Distance. The force of gravity can be calculated using the equation: Force = mass x acceleration due to gravity. Given that Alejandra weighs 225 newtons and the distance she climbs is 15 meters, we can plug these values into the formula to find the work done:

Work= (225 N) x (15 m) = 3375 Joules

Therefore, Alejandra does 3375 Joules of work against gravity when she climbs to the ledge at the top of the climbing wall.

Learn more about Work done against gravity here:

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A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?

Answers

75 km/hr, so in other words 75 km per hour.

7200 s / 60 = 120 min
120 min / 60 =2 hr
so 150 km / 2 hr = 75 km per hour

Answer:

The speed of the car is $\boxed{75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}}$ or $\boxed{20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The distance covered by the car is $150\,{\text{km}}$ .

The time taken by the car to cover the distance is $7200\,{\text{s}}$ .

Concept:

The speed of a car is the ratio of the distance covered by the car to the time taken by the car in covering the distance.

The expression for the speed of the car is:

$v = (d)/(t)$ …… (1)

Here, $v$ is the speed of the car, $d$ is the distance covered and $t$ is the time taken by the car.

Substitute the value of distance and time in the above expression.

$\begin{aligned}v &= \frac{{150\,{\text{km}}}}{{7200\,{\text{s}}\left( {\frac{{1\,{\text{hr}}}}{{60 * 60\,{\text{s}}}}} \right)}}\n&= \frac{{150}}{2}\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\n&= 75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\n\end{aligned}$

The speed of the car in terms of ${{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ can be expressed as.

$\begin{aligned}v&=75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}}{{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\left( {\frac{{1000\,{\text{m}}}}{{3600\,{\text{s}}}}} \right)\n&=75 *\frac{5}{{18}}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\n&= 20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\n\end{aligned}$

Thus, the speed of the car is $\boxed{75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}}$ or $\boxed{20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

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Answer Details:

Grade:Middle School

Chapter:Speed and distance

Subject:Physics

Keywords:Speed, velocity, distance, time, distance of 150 km, in 7200 s, speed of the car, ratio of distance, rate of change of distance.

A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. After a reaction time of 0.800 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

Answers

Answer:

t = 10.4 s

Explanation:

given,

velocity of the police car = 18 m/s

reaction time of policeman = 0.8 s

acceleration of the policeman to catch up the speeder = 5 m/s²

time taken to catch the speeder be t₁

distance traveled by the police and the speeder will be same

$D_p = D_s$

$u t_1 + (1)/(2)at_1^2 = u' t_1$

$18* t_1+ 0.5 * 5* t_1^2= 42* t_1$

$2.5 t_1^2 - 24 t_1 = 0$

$(2.5 t_1 - 24)t_1 = 0$

$t_1 = (24)/(2.5)$

t₁ = 9.6 s

total time taken by the policeman

= reaction time + t₁

= 0.8 + 9.6

t = 10.4 s

Total time taken by the policeman to catch the speeder is 10.4 s

Calculate the annual potential energy contained in the water that is held back by the dam. Assume an average height of the water, h of 100.0 m. Express the answer in GJ (at least three significant figures).

Answers

Answer:

The potential energy can be given as

E = mgh. m is mass, g = acceleration due to gravity = 9.8m/s, h is the heigh, given as 100.0m

E = m x 9.8 x 100 = (980m)J

E = (980m)/10^9GJ = (0.000000980m)GJ to 3 significant figures

Explanation:

Hydroelectric dams exploit storage of gravitational potential energy. A mass, m, raised a height, h against gravity, g = 9.8 m/s², is given a potential energy E = mgh. The result will be in Joules if the input is expressed in meters, kilograms, and seconds (MKS, or SI units).

A mass of 8 kg. moves at a rate of 14.3m/ sec. what is the KE developed by the mass

Answers

KE = 0.5 m v^2
= 0.5 * 8 * (14.3)^2
KE = 817.96 joule

The greater the___of the object, the greater the force needed to achieve the same change in motion.

Answers

Answer:mass

Explanation:

The momentum of a bald eagle in flight is calculated to be 345. The mass of the eagle is 5.0 kg. What is the magnitude of the velocity of the eagle?

Answers

p=mv
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