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Nady [450]
4 years ago
11

A car traveled at a constant velocity of 70 mph from noon to 2:00 pm. At 3:00 pm the velocity of the car was 80 mph; and finally

at 4:30 pm the velocity of the car was 40 mph.
Which statement accurately describes the acceleration of the car?
Physics
1 answer:
Paha777 [63]4 years ago
6 0

Answer:

Not sure which statements were give to you for this question, but the vehicle's acceleration from noon to 2:00 PM was <em>zero</em>, the vehicle had a <em>positive</em> acceleration between 2:00 pm and 3:00 pm, and the vehicle had a <em>negative</em> acceleration at 4:30 PM.

Explanation:

From 12:00 PM to 2:00 PM, the vehicle traveled at a constant velocity of 70 MPH, meaning there wasn't any change to the speed. The velocity <em>remained the same</em>. An hour later, the velocity of the vehicle <em>increased</em> to 80 MPH, and finally at 4:30 PM, the velocity of the car <em>decreased </em>and was at 40 MPH.

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An upright spring with a 96g mass on it is compressed 2 cm. When
Alexeev081 [22]

Answer:

I only know answer A and it's 2825.28 N/m, with rounding it's 2825.5

Explanation:

Use the m*g*h=1/2*k*x^2 equation

96*9.81*60=1/2*k*2^2

5650.56=2k

5650.56/2=2825.28N/m

8 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Help asap please I will give you 5stars
boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega

Hence, this is the required solution.

6 0
3 years ago
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d
zvonat [6]

Answer:

<u />D_l=d<u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

D=\frac{at^2}{2}

Where

Acceleration is given as

a=\frac{V_o}{2d}

And

Time

T=\frac{d}{v_0}

Therefore horizontal displacement towards the left is

D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}

<u />D_l=d<u />

5 0
3 years ago
Brainliest if correct Question 10 of 10
11Alexandr11 [23.1K]

Answer:

D thermal energy moves from the marshmallow to your fingers as you touch the marshmallow

Explanation:

Conduction is the movement of thermal energy without the actual movement of the particles of a body. They particles only collide with each other and transfer energy in the process while maintaining their mean positions.

5 0
2 years ago
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