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Alisiya [41]
3 years ago
9

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

ve period? .
Physics
1 answer:
Ann [662]3 years ago
8 0
We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
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The answer is number D
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3 years ago
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A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact w
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Answer:

Last option 3000N

Explanation:

4 0
3 years ago
A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a
uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}

Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

98\times4.2+85\times5.5=(98+85)\times v

v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0
3 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr
slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

E=\frac{1}{2} *L*I^{2}

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}

Then:

E'=4*E

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

3 0
3 years ago
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