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3 years ago

What is the symbol for a change in position? Ad p d Ap

Physics

1 answer:

Rudiy273 years ago

7 0

change in position of an object we use the

symbol ∆×\ delta×∆×for displacement, where ∆. means

"change" ∆ vector quantity with units of distances start

text ,d,I,s,t,a,n,c,e , end text.

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Calculate Vector component in Y if the hypotenuse is 32 and angle is 45

Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

$v_y = v \times sin(\theta)$

Substituting the values from the question gives;

$v_y = 32 \times sin(45^{\circ}) \approx 22.6$

The vector component in the y-axis, $v_y$ , is approximately 22.6.

8 0

3 years ago

A long point-object, mass = 1.0 kg, moves in a circular path at a radial distance = 0.5 m from the axis of rotation. What is the

Vinvika [58]

Answer: $0.25\ kg-m^2$

Explanation:

Given

mass of Point object $m=1 kg$

Distance $r=0.5 m$

Since mass is moving in circular path therefore every time mass is at distance of r from center .

Also Moment of Inertia tells about the distribution of mass over the given region with respect to center of mass.

Therefore $I=mr^2$

$I=1\times 0.5^2$

$I=0.25\ kg-m^2$

5 0

3 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

8 0

3 years ago

Read 2 more answers

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

3 years ago

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86

Vladimir [108]

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

3 0

3 years ago