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3 years ago

6

Find the mass of a golfball that accelerates at 18m/s2 when 8.3 N force is applied to it

1 answer:

joja [24]3 years ago

4 0

Answer:0.46kg

Explanation:

m=f/a

M=8.3/18

M=0.46kg

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A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2

Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass $m_1 = 10 g= 10^{-2} \ kg\\\\$

initial speed $u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\$

block mass $m_2 = 4\ Kg$

initial speed $v_2 =-4.2 \frac{m}{s}$

final speed $v_2= 0$

Let $v_1$ will be the bullet speed after collision:

throughout the consevation the linear moemuntum

$\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\$

$= 320 \frac{m}{s}$

The kinetic energy of the bullet in its emerges from the block

$k=\frac{1}{2} m_1 v_1^2$

$=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J$

3 0

3 years ago

IF YOU ANSWER ALL I WILL GIVE A BRAINLIEST... ONLY 3 QUESTIONS!

Law Incorporation [45]

Answer: a) electromagnetic waves

Explanation:

An electromagnetic wave begins when an electrically charged particle vibrates. This causes a vibrating electric field, which in turn creates a vibrating magnetic field. The two vibrating fields together form an electromagnetic wave.

Hope this helps:)

6 0

3 years ago

If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

8 0

3 years ago

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

3 years ago

A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

$\mu = \frac{m}{L}$

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

$\mu = \frac{0.09\ kg}{1\ m}$

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

7 0

3 years ago