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alukav5142 [94]

2 years ago

11

A bicycle and its rider have a combined mass of 80 kg and a speed of 6 m/s. What is the magnitude of the average force needed to

bring the bicycle and its rider to a stop in 4 seconds?

2 answers:

anygoal [31]2 years ago

6 0

Answer: 120N

Explanation:Please see attachment for explanation

lisabon 2012 [21]2 years ago

4 0

The magnitude is 24 meter

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If a 60-g object has a volume of 30 cm", what is its density?

alisha [4.7K]

Answer:

Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2

5 0

2 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

What are things that we can do to protect our climate for future generations?

Vilka [71]

<span>Reduce energy use.
Change the way you think about transportation. Walk or bike whenever possible.
Insulate your home. Insulate yourself and your home.
Make every drop count.
</span>Cool wash and hang to dry.
<span>Switch to "green power.
</span>Recycle.

6 0

2 years ago

A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont

AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as $dcos\theta$

Therefore, $W=Fdcos\theta$

Making F the subject of the formula then

$F=\frac {W}{dcos\theta}$ where $\theta$ is the angle of inclination. Substituting 190 J for W then 18 degrees for $\theta$ and 8 m for d then

$F=\frac {190}{8cos18^{\circ}}\approx 25N$

3 0

2 years ago