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3 years ago

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In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

m 0 mph to 6.7m/s in the same amount of time. What is the acceleration of the truck in m/s? *
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1 answer:

ANTONII [103]3 years ago

6 0

Is there a picture?

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An object is moving along a straight line at a

Dominik [7]

The answer for this question is 5 m

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2 years ago

If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v

soldier1979 [14.2K]

Answer:

$v = -32 t$

Explanation:

given,

s = 400- 16 t²

we know,

Velocity of an object is defined as the change in displacement per unit change in time.

velocity an also be return as

$v = \dfrac{ds}{dt}$

$v = \dfrac{d}{dt}(400-16t^2)$

$v= 0 -2\times 16 t$

$v = -32 t$

Hence, instantaneous velocity function given by $v = -32 t$

To calculate instantaneous velocity, you need to insert value of time.

ex, instantaneous velocity at t = 4 s

v = -32 x 4 = -128 m/s.

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3 years ago

Two forces act on an object. The first force has a magnitude of 15.0 N and is oriented 78.0 ° counterclockwise from the + x ‑axi

bezimeni [28]

Answer:

c

Explanation:

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3 years ago

A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

$\vec{F_B} = q(\vec{V}\times \vec{B})$

$F_B = q|v||B| sin\theta$

Here,

q = Charge

v = Velocity

B = Magnetic field

$\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}$

Our values are given as,

$\theta = 35.7\°$

$q = 22.5*10^{-9}C$

$B = 1.05*10^{-5}T$

$v = 3.11m/s$

Replacing,

$F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)$

$F_B = 4.224*10^{-13}N$

Therefore the size of the magnetic force acting on the bumble bee is $4.22*10^{-13}N$

3 0

3 years ago

A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

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3 years ago