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3 years ago

In which situation are both the speed and velocity of the car changing?

2 answers:

6 0

Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant..
Hope this helped.

NemiM [27]3 years ago

6 0

The answer to this question is:
In which situation are both the speed and velocity of the car changing?
A) The car turns left at 5
m
s
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If I made a mistake please feel free to tell me ^-^

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Two football players move in a straight line directly toward one another. Their equations of motion are as follows:

iren [92.7K]

The time when the two players will collide is 0.96 s.

The equation of motion of the two players is given as;

x1 = 0.1 m + (–3.9 m/s )t

x2 = –6.3 m + (2.8 m/s )t

The time when the two players collide, their displacement is equal or the difference in their position will be zero.

$X_1 - X_2 = 0\\\\0.1 - 3.9 t - (-6.3 + 2.8t) = 0\\\\0.1 -3.9t + 6.3 -2.8t = 0\\\\6.4 -6.7t = 0\\\\6.7t = 6.4\\\\t = \frac{6.4}{6.7} = 0.96 \ s$

Thus, the time when the two players will collide is 0.96 s.

Learn more here: brainly.com/question/18033352

6 0

2 years ago

In the movie Speed, a bus already moving faster than 55 mph will explode if the bus’ speed falls below 55 mph. Let say that the

kompoz [17]

Answer:

The bus moved after our girl Sandy passed out = S = 905.04 m

Explanation:

At t = 0 sec Vi = 80mph = 35.76 m/s

t = 30 sec Vf = 55mph = 24.58 m/s

According to the first equation of motion

Vf = Vi + at

24.587 = 35.76 + a(30)

a = - 0.373 m/s2 (negative sign shows deceleration)

Now for distance travelled in 30 sec we have

second equation of motion

S=Vit+1/2at2

S = 35.76 x 30 + 1/2 x -0.373 x 30 x 30

S = 905.04 m

7 0

3 years ago

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if

xxMikexx [17]

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

(b) The plates can be close together to 1.7 mm with this applied voltage

Explanation:

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

$E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ( $2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$ ) does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

b) To find how close together can the plates be with this applied voltage:

The formula would be,

$d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

$d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$

3 0

3 years ago

The mass of the Sun is 2x1030 kg, and the mass of Mars is 6.4x1023 kg. The distance from the Sun to Mars is 2.3X1011 m. Calculat

user100 [1]

Answer: force from sun to mars = $1.62*10^{19} N$ , force from mars to sun = $-1.62*10^{19} N$

Explanation: The force of attraction or repulsion between 2 object ( in this case the sun and mars ) in space is given by newton's law of gravitation which is given mathematically below as

$F =\frac{Gm_{s} m_{m} }{r^{2} }$

where $G = 6.67 * 10^{-11} , m_{s} = 2*10^{30}kg m_{m} = 6.4 *10^{23} kg\\r = 2.3 *10^{11} m$

G = gravitational constant, $m_{m}$ = mass of mars, $m_{s}$ = mass of sun

$F = \frac{6.64 * 10 ^{-11}* 2 * 10^{30} *6.4*10^{23}} {2.3*10^{11} * 2.3* 10^{11} } =\frac{85.376 *10^{42} }{5.25 *10^{22} } \\\\F = 1.62 * 10 ^{19} N\\$

the mars also exerts the same magnitude of force above on the sun but in the opposite direction, thus the force mars exerts on the sun is $-1.62*10^{19} N$

8 0

3 years ago

Prove that hoop stress is twice the longitudinal stress in a cylindrical pressure vessel.

Iteru [2.4K]

Answer:

Proof in explanation.

Explanation:

Consider a thin cylinder, whose thickness to diameter ration is less than 1/20, the hoop stress can be derived as follows:

Let,

L = length of cylinder

d = internal diameter of cylinder

t = thickness of wall of cylinder

P = internal pressure

σH = Hoop Stress

σL = Longitudinal Stress

Total force on half-cylinder owing to internal pressure = P x Projected Area = P x dL (Refer fig 9.1)

Total resisting force owing to hoop stress setup in walls = 2 σH L t

Therefore,

P d L = 2 σH L t

σH = Pd/2t _____________ eqn (1)

Now, for longitudinal stress:

Total force on end of cylinder owing to internal pressure = P x Projected Area = P x πd²/4

Area resisting this force = π d t (Refer fig 9.2)

Longitudinal Stress = Force/Area

σL = (Pπd²/4)/(πdt)

σL = Pd/4t ____________ eqn (2)

Dividing eqn (1) by eqn (2)

σH/σL = 2

σH = 2 σL (Hence, Proved)

6 0

3 years ago