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3 years ago

A car accelerated from rest in 3.07s. During this time, the car went 46m forward. What was its acceleration?

Physics

1 answer:

kari74 [83]3 years ago

8 0

Vi=0
T=3.07s
D=46m
A=?
A=2(d)/t ^2
2(46)/3.07^2=9.8m/s^2

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A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

WITCHER [35]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

8 0

3 years ago

A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at

4vir4ik [10]

Answer:

0.541 nm

Explanation:

The condition for maxima is,

$dsin\theta=m\lambda$

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, $\lambda$ is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

$\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m$

And the order of maxima is $m=1$ .

And the angle at which first order maxima occur is, $\theta=33^{\circ}$ .

Put these values in maxima condition while solving for d.

$d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm$

Therefore, the slit separation is 0.541 nm.

7 0

3 years ago

Coulomb's law is expressed mathematically as

nataly862011 [7]

Force between two charges =

( 1/4πε₀ ) · (Charge #1) · (Charge #2) / (Distance between them)²

in the direction away from each other.

In other words, if the force is positive, the charges are repelling.
If the force is negative, the charges are attracting.

4 0

3 years ago

The mass of the basketball shown below is 4 times the mass of the baseball.

igomit [66]

Answer:

<h3>40.0 Joules. None of the answers is correct</h3>

Explanation:

Kinetic energy of an object is expressed as;

KE = 1/2mv² where;

m is the mass of the object

V is the velocity of the object.

Let Ma be the mass of the basketball and Mb be the mass of baseball.

If the mass of the basketball shown below is 4 times the mass of the baseball, then Ma = 4Mb

KE of the basket ball = 1/2MaVa²

Va² = 2(KE)a/Ma ...... 1

KE of the baseball = 1/2MbVb²

Vb² = 2(KE)b/Mb .......... 2

Since their velocities are the same, hence Va²= Vb²

2(KE)a/Ma = 2(KE)b/Mb

Substituting Ma = 4Mb into the resulting expression.

2(KE)a/4Mb = 2(KE)b/Mb

2(KE)a/4 = 2(KE)b

Given Kinetic energy of baseball (KE)b = 10.0J, the expression becomes;

2(KE)a/4 = 2(10)

2(KE)a = 4*20

2(KE)a = 80

(KE)a = 80/2

(KE)a = 40.0J

Hence the kinetic energy of the basketball is 40.0Joules

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3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago