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3 years ago

8

Please help answer both of these !! I’m in a rush :(((

1 answer:

Leno4ka [110]3 years ago

7 0

Answer: search it on browser

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A 3.50 kg block is pushed along a horizontal floor by a force of magnitude 15.0 N at an angle θ = 30.0° with the horizontal. The

Crank

Answer :

The frictional force on the block from the floor and the block's acceleration are 10.45 N and 0.73 m/s².

Explanation :

Given that,

Mass of block = 3.50

Angle = 30°

Force = 15.0 N

Coefficient of kinetic friction = 0.250

We need to calculate the frictional force

Using formula of frictional force

$F_{k}=\mu N$

$F_{k}=\mu (F\sin\theta+mg)$

$F_{k}=0.250\times(15\times\sin30^{\circ}+3.50\times9.8)$

$F_{k}=0.250\times41.8$

$F_{k}=10.45\ N$

(II). We need to calculate the block's acceleration

Using newton's second law of motion

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{F\cos\theta-F_{k}}{m}$

$a=\dfrac{15.0\cos30^{\circ}-10.45}{3.50}$

$a=0.73\ m/s^2$

Hence, The frictional force on the block from the floor and the block's acceleration are 10.45 N and 0.73 m/s².

8 0

3 years ago

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1

SCORPION-xisa [38]

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

5 0

2 years ago

The cycle of the moon through its phases or the synodic month is___.

ikadub [295]

It is 29 and a half days long

3 0

3 years ago

Read 2 more answers

Which simple machine is NOT correctly matched with an appropriate task for its use?

Tcecarenko [31]

The simple machine that is not correctly matched with its appropriate task is the inclined plane because there is no such big ramp that is as high as 1 storey building, the appropriate task would be Lifting a heavy box and moving it across a room. and for the pulley : <span>Moving a heavy box up to the second floor of a building.</span>

6 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago