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NikAS [45]
4 years ago
11

Marissaâs car accelerates uniformly at a rate of +2.60 m/s². How long does it take for Marissaâs car to accelerate from a speed

of 24.6 m/s to a speed of 26.8 m/s?
Physics
1 answer:
DIA [1.3K]4 years ago
4 0

Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, a=+2.6\ m/s^2

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

t=\dfrac{v-u}{a}

t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.                                    

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A stationary block resting on the ground is pulled up with a tension force of 100N, but does not leave the ground.
tekilochka [14]

Answer:

The normal force the ground exerts on the block, F = -300 N

Explanation:

Given data,

The block pulled up with a tension force, T = 100 N

The weight of the block, W = 300 N

The weight of the block is due to the force of attraction of gravitation.

The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.

The weight of the block,

                                  W = mg

                                          300 N

The normal force the ground exerts on the block,

                                  F = - mg

                                    = - 300 N

Hence, the normal force the ground exerts on the block, F = -300 N

6 0
3 years ago
A moving walkway has a speed of .4 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of 3
SVEN [57.7K]

Explanation:

Vnet = 0.4m/s - 3.0m/s = - 2.6m/s

so the man walks 2.6m/s towards the west

8 0
3 years ago
Resistance training should be performed every day for maximum conditioning benefits.
marysya [2.9K]

Answer:

The answer is True

7 0
3 years ago
Read 2 more answers
What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 se
shusha [124]

Answer:

Acceleration, a=\dfrac{1}{8}(-i+9j)\ m/s^2

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}    

a=\dfrac{(-i+9j)}{8\ s}\ m/s^2    

or

a=\dfrac{1}{8}(-i+9j)\ m/s^2

Hence, the value of acceleration vector is solved.

4 0
4 years ago
The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

#SPJ9

4 0
1 year ago
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