All categories

3 years ago

Can someone help me?!!!!!

Physics

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

7 units, north

Explanation:

This is a problem of vector subtraction. We have:

- Vector A: magnitude 2 units, direction to the north

- Vector B: magnitude 5 units, direction to the south

If we take the north as positive direction, we can write

$A = +2\\B=-5$

Since we want to find $A-B$ (vector subtraction), we have to change the sign of B, so we find:

$A-B=+2-(-5))=+2+5=+7$

And the positive sign means the direction is north.

You might be interested in

Calculate the magnitude of the normal force on a 25.2 kg block in the following circumstances. (Enter your answers in N.) HINT (

irga5000 [103]

Answer:

when the body is resting N = 246.96 N

when the body is resting on a tilted surface N = 212.12 N.

when the body is in a elevator N = 317.036 N

Explanation:

when the block is resting on a stationary surface the normal force is balanced by the weight of the body.

weight of the body = mg = 25.2×9.8 = 246.96 N

therefore normal force = 246.98 N.

when the block is resting on a tilted surface the normal force will be balanced by the $\cos \theta$ component of the weight where Ф is the angle of inclination.

therefore N = mg $\cos \Phi$

N= 212.12 N.

when the block is resting on a elevator that is accelerated upward the normal force will be the sum of weight and force due to acceleration ma

therefore N = 246.98 + 25.2×2.78

N = 317.036 N

5 0

4 years ago

A closed box is filled with dry ice at a temperature of -87.1°C, while the outside temperature is 17.6°C. The box is cubical, me

natta225 [31]

Answer:

K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹

Explanation:

Rate of flow of heat through a material is given by the following expression

$\frac{Q}{t} =\frac{KA\delta T}{d}$

where Q is amount of heat flowing in time t through area A and a medium of thickness d having two faces at temperature difference δT . K is thermal conductivity of the medium .

Here Q = 3.34 x 10⁶/6 , t = 24 x 60 x 60 = 86400 s , A = .332 X .332 = .0110224 m² , δT = 104.7

Put these values here

$\frac{3.34\times10^6}{6\times86400}= \frac{k\times.011224\times104.7}{4.41\times10^{-2}}$

$K=\frac{3.34\times4.41\times10^4}{6\times86400\times.011224\times104.7}$

K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹

4 0

3 years ago

Which of the following statements must be true if the net force on an object is zero? Choose all that apply.

Black_prince [1.1K]

Answer:

a) The object must have constant velocity.

d) The object must have zero acceleration.

Explanation:

We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:

$F = ma$

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the net force on the object is zero:

F = 0

This means that the acceleration of the object is also zero, according to the previous equation:

a = 0

So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

Which means that $\Delta v=0$ , so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:

b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero

c)The object must be at the origin. --> false, since the object can be in motion

5 0

3 years ago

A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at

ivanzaharov [21]

(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

4 0

4 years ago

Can anyone help me I will give you 30 points

Aleonysh [2.5K]

Answer:

1) ELECTRO MAGNET

2) B

3) -AMOUNT OF ELECTRICITY OF THE POWER SOURCE

- AMOUNT OF COIL WRAPPED UP TO THE NAIK

-CONDUCTIVITY OF THE MEDIUM

3 0

4 years ago