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3 years ago

Calculate the man’s mass. (Use PE = m × g × h, where g = 9.8 N/kg.)

Physics

1 answer:

vodomira [7]3 years ago

7 0

Answer:

mass=56.12kg

Explanation:

PE=mgh

4620=m×9.8×8.4

make m subject of the formula...

m =4620/(9.8×8.4)

m=4620/82.32

m=56.12kg

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A mass of 148g stretches a spring 6cm. The mass is set in motion from its equlibrium position with a downward velocity of 10cm/s

zalisa [80]

Answer:

$u(t)=0.78sin12.78t$

Explanation:

We are given that

Mass,m=148 g

Length,L=6 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}=\sqrt{\frac{980}{6}}=12.78 rad/s$

Where g= $980 cm/s^2$

$u(t)=Acos12.78 t+Bsin 12.78t$

u(0)=0

Substitute the value

$A=0$

$u'(t)=-12.78Asin12.78t+12.78 Bcos12.78 t$

Substitute u'(0)=10

$12.78B=10$

$B=\frac{10}{12.78}=0.78$

Substitute the values

$u(t)=0.78sin12.78t$

7 0

2 years ago

Two small insulating spheres with radius 3.50×10^−2m are separated by a large center-to-center distance of

Liula [17]

Answer:

Part (i) the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (ii) the direction of the electric field is towards the negative charge

Explanation:

Given;

+q = 3.93.90μC, r = 3.50×10⁻²m

-q = −2.40μC, r = 3.50×10⁻²m

magnitude of the electric field is experienced, midway between the spheres at a distance r, r = ¹/₂ × 0.51 = 0.255 m

Electric field due to point charge is given as;

$E = \frac{F}{q} = \frac{Kq^2}{qr^2} = \frac{kq}{r^2}$

K is coulomb's constant = 8.99 x 10⁹ Nm²/C²

The positive charge on positive x-axis and the negative charge is on negative x-axis.

part (a)

The electric field due to positive charge; +q = 3.93.90μC

$E_+ = \frac{kq}{r^2}\\\\E_+ = \frac{8.99 X10^9*3.9X0^{-6}}{0.255^2}\\\\E_+= 5.3919 X10^5\frac{N}{C}$

The electric field due to negative charge; -q = −2.40μC

$E_- = \frac{kq}{r^2}\\\\E_- = \frac{8.99 X10^9*2.4X0^{-6}}{0.255^2}\\\\E_-= 3.3181 X10^5\frac{N}{C}$

From superimposition theorem

The magnitude of the electric field is;

E = E₊ + E₋

E = (5.3919 × 10⁵ + 3.3181 × 10⁵) N/C

E = 8.71 × 10⁵ N/C

Therefore, the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (b)

The direction of the electric field is towards the negative charge.

7 0

3 years ago

If you were to stir salt in water until the salt disappeared, which is the solute?

emmainna [20.7K]

Salt is the solute in that situation

6 0

2 years ago

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My question is An object may be acted on by several forces. What name is given to the single force that has the same effect as t

wariber [46]

Newtons Law of motion

6 0

3 years ago

Read 2 more answers

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

3 years ago