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3 years ago

8

Which statement is true of all matter?

1 answer:

Fittoniya [83]3 years ago

8 0

Answer:

D. It must have mass and volume

Explanation:

In science, matter is referred to as any substance that has weight and occupies space. This means that the substance must have a MASS of its own when weighed and also a VOLUME.

Matter include elements, molecules, humans, etc. In fact, almost every substance on Earth is considered MATTER. Therefore, the fact that a substance must "have mass and volume" is true for all matter.

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If the sum of the external forces on an object is zero, then the sum of the external torques on it

Zarrin [17]

Answer:

True.

Explanation:

If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.

The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

Hence, the given statement is true.

5 0

3 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

3 years ago

Read 2 more answers

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago

Một dây nhôm dài 10 m khi ở 25 độ C. Biết khi nhiệt độ tăng thêm 1 độ C thì chiều dài 1m dây nhôm sẽ tăng thêm 0,024mm.

kondor19780726 [428]

Answer:

??????? what language is this. um A then

5 0

3 years ago

An electric motor converts electrical energy into (blank) energy

Delicious77 [7]

Answer:

Mechanical

Explanation:

5 0

3 years ago