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KengaRu [80]
3 years ago
15

How does an object's position and velocity change as the object accelerates?”

Physics
1 answer:
taurus [48]3 years ago
7 0

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

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Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

Average speed of Prav,

v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0
3 years ago
In which of the following elevator situations would the acceleration be positive. Select TWO answers
kap26 [50]

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

4 0
3 years ago
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
What will happen to the 0.1 N force if one charges is increased by a factor of 3?
AleksandrR [38]

Answer:

Explanation:

Force between two charges is given by the following expression

F = \frac{KQ_1Q_2}{d^2}  Q₁ and Q₂ are two charges and d is distance between two.

.1 = \frac{KQ_1Q_2}{d^2}

If Q₁ becomes three times , force will become 3 times . Hence force becomes .3 N in the first case.

Force F = .3 N

If charge becomes one fourth , force also becomes one fourth .

F= \frac{.1}{4}

= .025 N.

5 0
3 years ago
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