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3 years ago

what is the acceleration of a 2,000- kilogram truck if a force of 4,200. N is used to make it start moving forward?

1 answer:

5 0

Newton's 2nd law of motion: Force = (mass) x (acceleration)

Divide each side by (mass) : Acceleration = (force) / (mass)

Acceleration of the truck = (4,200 N) / (2,000 kg) = 2.1 m/s²

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A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f

Sloan [31]

A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.

8 0

3 years ago

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0

3 years ago

A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

$\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s$

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0

3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

4 years ago