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rosijanka [135]
3 years ago
10

Dave is moving 3 m/s when he crashes his bike into a wall, which stops him in 0.6 seconds. If Dave and his bike have a mass of 9

0 kg,
what was the force applied by the wall?
Physics
1 answer:
Arisa [49]3 years ago
7 0
Force = change of momentum / time taken
Force = (90x3)/0.6
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3 years ago
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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0
2 years ago
The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

E = 1.655 x 10⁷ N/C

4 0
3 years ago
An object has a kinetic energy of 275 j and the momentum of magnitude of 25.0 kg m/s. find the speed and mass of the object
lyudmila [28]
<span>Recall formula for Kinetic energy is: KE = 1/2mv^2, where KE = 275J and momentum (which is 25.0 kg m/s) = m*v Therefore substitute for KE and mv in the equation above to get speed => 275 = 0.5 * 25 * v v = 275/12.5 v = 22m/s to get mass m, recall momentum = m*v => 25= 22*m m= 25/22 = 1.3663kg</span>
6 0
3 years ago
Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit
ankoles [38]

Answer:

The acceleration of man 1 and 2 is 3.125\ m/s^2 and 4.167\ m/s^2.

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2

Let a₂ is acceleration of man 1,

F = m₂a₂

a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2

So, the acceleration of man 1 and 2 is 3.125\ m/s^2 and 4.167\ m/s^2.

7 0
2 years ago
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