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tia_tia [17]
3 years ago
10

An intercontinental ballistic missile goes from rest to a speed of 6.50 km/s in 60.0 s. In multiples of g, what is its accelerat

ion?
Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The acceleration is 11.1g\ m/s^2

Explanation:

Given that,

Speed v= 6.50\ km/s=6.5\times10^{-3}\ m/s

Time t = 60.0 sec

We need to calculate the acceleration

Using formula off acceleration

a = \dfrac{\Delta v}{t}

a=\dfrac{v_{f}-v_{i}}{t}

We know that,

Missile goes from rest

So, Initial velocity =0

Put the value into the formula

a =\dfrac{6.50\times10^{3}}{60.0}

a=108.33\ m/s^2

On right hand side multiplying and dividing by g = 9.8m/s²

a=108.33\times\dfrac{g}{g}

Put the value of g

a = \dfrac{108.33}{9.8}g\ m/s^2

a = 11.1g\ m/s^2

Hence, The acceleration is 11.1g\ m/s^2

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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

4 0
3 years ago
When can a theory be modified if a new type of technology allows for new observations that raise new questions?
olga nikolaevna [1]

Answer:

id say B, but not sure

Explanation:

4 0
3 years ago
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- Name two elements that have the SAME (rounded) atomic mass:
katovenus [111]

Answer:

Magnesium atomic no. = 24,25,26. These are the two elements which have same atomic no

Explanation:

3 0
2 years ago
Help, anyone?? please:/
NeTakaya
A,E,C,B,D is the order
5 0
3 years ago
The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?
anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
where
\rho is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
\rho' =  \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4  \frac{\pi r^2 R}{L}   = 4 \rho
Therefore, the new resistivity must be 4 times the original one.
5 0
3 years ago
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