All categories

3 years ago

Help needed. The mass of an object times its velocity is the blank of the object ?

Physics

2 answers:

Kryger [21]3 years ago

4 0

Momentum.

Remember the formula p=mv?

Svetach [21]3 years ago

4 0

Momentum of an object is the product of mass and velocity.

<u>Explanation:</u>

Momentum is a quantity of motion of a moving body. Let us indicate mass of the object by m, and velocity of the object by v. It is a well-known concept that the mass of the object irrespective of its velocity or motion always remains constant.

And the velocity of the object might change when the distance travelled by the object or the time is said to vary. So the Momentum is required to force or speed the mass of an object.

You might be interested in

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

What is the momentum of a 0.2kg ball with a speed of 5 m/s

Eddi Din [679]

Momentum

= mass x velocity

= 0.2 x 5

= 1 kg m/s

3 0

3 years ago

A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

$F = 14 lbf$

$F = 6.35 \times 9.8 N$

$F = 62.3 N$

stretch in the spring is given as

$x = 4 in = 0.1016 m$

now we will have

$F = kx$

$62.3 = k(0.1016)$

$k = 613.125 N/m$

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

$W = \frac{1}{2}kx^2$

$W = \frac{1}{2}(613.125)(0.254)^2$

$W = 19.8 J$

7 0

3 years ago

What is the definition of acceleration

PtichkaEL [24]

Acceleration is the way the motion is changing.

4 0

3 years ago

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an

astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution= $T_b=101^{\circ}$ C

Boiling point water=100 degree Celsius

$K_f=1.86K/m$

$K_b=0.512 K/m$

$\Delta T_b=T-T_0$

Where $T$ =Boiling point of solution

$T_0=$ Boiling point of pure solvent

$\Delta T_b=101-100=1^{\circ}$ C

$\Delta T_b=k_bm$

Using the formula

$1=0.512\times m$

Molality, $m=\frac{1}{0.512}$ m

$\Delta T_f=k_fm$

Using the formula

$\Delta T_f=\frac{1}{0.512}\times 1.86$

$\Delta T_f=3.63 C$

We know that

$\Delta T_f=T_0-T_1$

Where $T_0$ =Freezing point of solvent

$T_1=$ Freezing point of solution

Using the formula

$3.63=0-T_1$

Freezing point of water=0 degree Celsius

$T_1=0-3.63=-3.63 C$

Hence, the freezing point of solution=-3.63 degree Celsius

7 0

3 years ago