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kogti [31]
3 years ago
14

1. A bicyclist travels for 1.5 hours at an average speed of 23 km/h. How far does the bicyclist travel

Physics
2 answers:
allsm [11]3 years ago
8 0

Explanation:

23 km/hr x 1.5 hrs = 34.5 km

34.5 km = 34500 meters

SOVA2 [1]3 years ago
5 0
In this case your finding d or distance. So you would have to multiply the time by speed. The time in this case is 1.5hrs, the speed in 34.5km/h. So multiply and you’ll get 34.5m
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Find τf, the torque about point p due to the force applied by the achilles' tendon.
Luba_88 [7]
The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon

If there are given values, substitute in the equation and solve for the torque.</span>
3 0
3 years ago
The magnitude of the momentum of an object is 64 kilogram meters per second. If the magnitude of te velocity is doubled, the mag
dsp73

Answer:

128 \frac{kg*m}{s}

Explanation:

P=m*v\\if double velocity:\\P=m*2v=2m*v\\

7 0
3 years ago
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
Why is it very difficult for astronomers to find objects like planets and asteroids?
mezya [45]
It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location
3 0
3 years ago
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