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Tresset [83]
3 years ago
6

A projectile is launched horizontally at a speed of 45.0 m/s from a

Physics
1 answer:
Free_Kalibri [48]3 years ago
8 0

Answer:

answer will be c: 1.5 seconds :)

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What is the acceleration of a 1000 kg car subject to a 550 N net force
Xelga [282]

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

PLEASE MARK ME AS BRAINLIEST!!!

4 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium
ss7ja [257]

Answer:

The number of atoms are 1.86\times10^{8}.

Explanation:

Given that,

Diameter D = 1.40\times10^{2}\ pm

D=1.40\times10^{2}\times10^{-12}\ m

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}

Number\of\atoms =1.86\times10^{8}

Hence, The number of atoms are 1.86\times10^{8}.

8 0
3 years ago
A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and
Airida [17]

Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

Combined moment of inertia of stool,student and bricks =6\ kg.m^2

Now student pull off his hands so as to increase its speed to suppose N_2 rev/s

\omega _2=2\pi N_2  

After Pulling off hands so final moment of inertia is

I_2=2\ kg-m^2

Conserving angular momentum  as no external torque is applied

I_1\omega _1=I_2\omega _2

6\times 6.284=2\times \omega _2

\omega _2=18.85\ rad/s

N_2=3 rev/s

7 0
3 years ago
A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a
Slav-nsk [51]

For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

6 0
3 years ago
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