The number of planets, besides the earth, that are visible to the unaided eye is

A cyclist overcomes a resistive force of 30 N in order to cycle 30 m. It takes her 6 seconds to cycle this distance. Calculate t

zmey [24]

Answer:

Power = 15[W]

Explanation:

This problem can be solved using the work definition.

Work is equal to the product of the force by the distance, for this problem we have:

F = force = 30 [N]

d = distance = 30 [m]

w = work = F * d = 30*30 = 900 [J], "units in joules"

The power is defined as the work done in an interval of time.

P = w / t

where:

t = time [s]

therefore

P = w / t

P = 900 / 6

P = 150 [W] "units in watts"

6 0

3 years ago

A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

shusha [124]

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

$y=y_0+v_ot +\dfrac{1}{2}gt^2$

Putting all the values,

$0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s$

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

$d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m$

3 0

3 years ago

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Kamila [148]

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

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

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

${30}^{2} - {0}^{2} = 2 \times 10 \times h$

$= > 20h = 900$

$= > h = \frac{900}{20} = 45$

5 0

3 years ago

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

5 0

2 years ago

A psychologist claiming that a client's personal experiences and viewpoint influence behavior more than events in reality is usi

vodka [1.7K]

A psychologist who would claim that a client's personal experience and viewpoint influence behavior more than events in reality would probably use cognitive psychology mixed with developmental aspects to explain the behavior and personality of a person.

5 0

2 years ago

Read 2 more answers