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Cerrena [4.2K]
1 year ago
6

A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per seco

nd. the magnitude of the impulse applied to stop the cart is
Physics
2 answers:
ss7ja [257]1 year ago
6 0
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
 The magnitude of the impulse applied to stop the cart is
 I = FpΔt = (10N) * (5s) = 50 N.s
inn [45]1 year ago
4 0

Answer:The magnitude of the impulse applied to stop the cart is 50 N sec.

Explanation:

Force applied on the cart = 10 N

Interval of time till force applied ,Time= 5 second

Impulse=Force\times Time=10 N\times 5 seconds= 50 N seconds

The magnitude of the impulse applied to stop the cart is 50 N sec.

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A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.
kondaur [170]
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3 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
1 year ago
An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the
laila [671]

Answer:

a)  V = 1.866 10² V ,  b)   V = 3.424 10⁵ V , c)   v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

           v = 0.027 c

           v = 0.027 3 10⁸

           v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

           Em₀ = U = q V

final point. Electrons with maximum speed

          Em_f = K = ½ m v2

          Em₀ = Em_{f}

          e V = ½ m v²

          V = ½ m v² / e

let's calculate

          V = ½  9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

          V = 1.866 10² V

           V = 1866 V

         

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

          V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

           V = 3.424 10⁵ V

           V = 342402 V

c)   this potential difference should give the protons the same speed as the electrons

             v = 8.1 10⁶ m / s

5 0
2 years ago
How high above the earth's surface is g reduced to 8.80m/^2?
Sladkaya [172]
Gravity changes as the altitude change.<span> The gravitational force is proportional to 1/R2,  where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get 

x = 345.85 km above the earths surface
</span>

<span>Hope my answer would be a great help for you.    
If </span>you have more questions feel free to ask here at Brainly.

<span> </span>


6 0
2 years ago
Read 2 more answers
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