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3 years ago

Based on the data given, in what direction will the car accelerate?

Physics

1 answer:

skad [1K]3 years ago

7 0

Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

(487-left + 632-right) - (632-right - 487-right) = 145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.

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Find the velocity (in m/s) of a proton that has a momentum of 4.96 X 10^-19 kg.m/s.

kramer

Answer:

The velocity of the proton is $2.965*10^{8} \frac{m}{s}$

Explanation:

The momentum of a particle is defined as the product of its mass by its velocity and we can calculate it using the following formula:

p=m*v Equation (1)

Where:

p: Is the momentum in kg*m/s

m: Is mass of the particle in kg

v: Is the velocity of the particle in m/s

Data known:

m= 1,6726 × 10^–27 kg : mass of the proton

p= 4.96 X 10^-19 kg.m/s.

We replace this data in the Equation (1):

$4,96*10^{-19} =1.6726*10^{-27} *v$ $v=\frac{4.9*10^{-19} }{1.6726*10^{-27} }$

$v=(\frac{4.96}{1.6726} )*(10^{27-19} )$

$v=2,965*10^{8} m/s$

Answer: The velocity of the proton is $2.965*10^{8} \frac{m}{s}$

3 0

3 years ago

Which of these properties change if the size of an object changes?

german

The correct answer is
<span>Density

In fact, the density of an object is the ratio between its mass m and its volume V:
</span> $d= \frac{m}{V}$
<span>if the size of the object changes, its volume V changes as well, therefore the density d must change.</span>

7 0

3 years ago

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

Gelneren [198K]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

3 years ago

Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.

Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

$f_n = nf_1$

where

n is the number of the harmonic

$f_1$ is the fundamental frequency

Here we have:

$f_3=512 Hz$ is the frequency of the 3rd harmonic

So the fundamental frequency is

$f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz$

And so, the frequency of the 2nd harmonic is:

$f_2=2f_1=2(170.7)=341.3 Hz$

3 0

3 years ago

A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se

Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²

d) θ = 52 rad

Explanation:

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

Tangential acceleration is calculated as follows:

at = α*R Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (3)

θ= ω₀*t + (1/2)*α*t² 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 speed ( rad/s)

ωf : final angular speed ( rad/s)

R : radius of the circular path (m)

at: tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m : radius of the disk

t₀=0 , ω₀ = 5 rev/s

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s =10π rad/s = 31.42 rad/s

ωf = 2*(2π)rad/s =4π rad/s = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0

3 years ago