An athlete of mass 70kg runs at a velocity of 6ms5-1 calculate the kinetic energy of the athlete.

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

$\frac{10N}{50kg}=a$

A newton is just $\frac{kg·m}{s^{2}}$

$a=\frac{\frac{10kg·m}{s^{2}}}{50kg}$

The s^2 and 50 kg you multiply

$a=\frac{10kg·m}{50kg·s^{2}}$

The kg's cancel and 10/50 is 1/5

$\frac{1}{5}·\frac{m}{s^{2}}$

So the acceleration is 1/5 m/s^2

3 0

3 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago

a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg

IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

$M = \frac{q}{p}$

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

$4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m$

Now using thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\$

6 0

3 years ago

The velocity of an object is positive and steadily increasing. Which of the following graphs represents how the acceleration of

stiv31 [10]

If an object's velocity is steadily increasing it means that the acceleration is constant at a certain value.

Choice A shows an acceleration of zero which would only be true if the object was not moving or if its velocity was not changing.

Choice B gives us a graph showing acceleration increasing over time and is therefore incorrect.

Choice C is correct because the acceleration is constant. Steadily increasing tells us that the acceleration is fixed at a certain value.

Choice D is incorrect an represents a constant negative acceleration. This would be the case if the object was steadily decreasing in velocity.

4 0

3 years ago

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

3 years ago

An athlete of mass 70kg runs at a velocity of 6ms5-1 calculate the kinetic energy of the athlete.​

An athlete of mass 70kg runs at a velocity of 6ms5-1 calculate the kinetic energy of the athlete.